Unit 8 — End of Topic Tests (reviewer copy: all keys and model answers visible)
End of Topic Test: Mole ratios from balanced equations & Mass-to-mass calculations (u08-t1 · 5 forms × 23 items · 115 total)
David's notes for James:
PRECISION-ALTERNATE CONTRACT (sol authoritative ruling, same class as U8-G2): on every numeric item with precisionAssessed=true, acceptedAlternates now byte-match the graded display up to a unit suffix or thousands-commas ('1,839.6' at B#9/E#9 kept). Bare-integer / short forms ('15', '6', '4 mol', ...) were REMOVED on A#9/A#11/A#12/B#11/C#11/D#8/D#11/E#11 — your grader must not re-accept forms that violate the stem's demanded precision.
REALISM REPLACEMENTS (sol loop-1): three combustion-product instances were chemically wrong and rebuilt — 4Na+O₂→2Na₂O (Na favors the peroxide; glass Na₂O comes from soda ash) ⇒ Kroll refining TiCl₄+2Mg→Ti+2MgCl₂ at A#19; 4K+O₂→2K₂O (K favors the superoxide) ⇒ calciothermic V₂O₅+5Ca→2V+5CaO at B#20; 2Co+O₂→2CoO (Co favors Co₃O₄) ⇒ direct nitridation 2Al+N₂→2AlN at D#14. All replacement constants are exact on the canonical 1-dp table; A#19/B#20 use only enrolled a8 atomic masses.
B#7 INSTANCE NOTE: the seed-derived 2N₂O→2N₂+O₂ ratio item was replaced with C₃H₈+5O₂→3CO₂+4H₂O asked O₂:C₃H₈ (5:1) because in the N₂O equation subscript-based reasoning reproduces both the reversal (1:2) and wrong-partner (2:2) values, so no wrong-answer feedback could uniquely name the student's error. In the propane instance every anticipated wrong (1:5 / 5:3 / 1:1) has exactly one producing mistake.
TYPED-RATIO LENIENCY (recorded, deliberate): the text-entry ratio contracts accept 'a : b' and 'a to b' alongside the requested colon form — the measured skill is the coefficient pair in the asked order, not the delimiter. sol left this as a minor note on C#2/C#7; consistent bank-wide (B#7 too).
C#14 DIAGNOSTICS: the wrong-value diagnostics 0.21 and 0.85 are the exact 2-dp error values; a student answering at the requested 1 dp types 0.2/0.9, which still lands inside the 0.05 matching tolerance. Labels and feedback verified correct — left as-is, flag if your matcher uses exact string compare.
GATE_CONFIG ADJUDICATIONS (unchanged through QC, full rationale inside gate_config.json + fix_log.md): a8-taught-constants family (42 enrolled 1-dp masses that recur by the STO parenthetical convention; every occurrence context-verified C1/C2/C3 and honesty-checked against the canonical table), equation-notation gram class (held by the bank-wide identical-equation guard — no two slots share a full equation), spec-verified task idioms + the ratio-colon convention phrase.
V-T12 TYPED-WORK CONVENTION (stage-3 ruling, carried): FRQ model answers and rubrics reference substances by worded names where students must type them — nothing typed requires sub/superscripts.
FRQ COMPOSITION DEVIATION (stage-1 ruling, flagged then, restating here): frq_per_form=3 OVERRIDES the seed file's frq_per_form=1 baseline per QB-v2 §4 (stoichiometry = calculation-heavy, three distinct taught procedure families: mole-ratio both roles / partial chains / full mass-to-mass chain + inspection). 23 items/form total.
FRQ delivery: your `prod` draw-on-paper class is delivered as a typed diagram-PLAN task (state kinds/recipes/counts/spacing; rubric grades the same observables). Confirm this mapping (pending from the U01-G1 round-trip).
full notes file
# u08-t1 — notes/queries from David's side (U8-G1)
1. PRECISION-ALTERNATE CONTRACT (sol authoritative ruling, same class as U8-G2): on every numeric item with precisionAssessed=true, acceptedAlternates now byte-match the graded display up to a unit suffix or thousands-commas ('1,839.6' at B#9/E#9 kept). Bare-integer / short forms ('15', '6', '4 mol', ...) were REMOVED on A#9/A#11/A#12/B#11/C#11/D#8/D#11/E#11 — your grader must not re-accept forms that violate the stem's demanded precision.
2. REALISM REPLACEMENTS (sol loop-1): three combustion-product instances were chemically wrong and rebuilt — 4Na+O₂→2Na₂O (Na favors the peroxide; glass Na₂O comes from soda ash) ⇒ Kroll refining TiCl₄+2Mg→Ti+2MgCl₂ at A#19; 4K+O₂→2K₂O (K favors the superoxide) ⇒ calciothermic V₂O₅+5Ca→2V+5CaO at B#20; 2Co+O₂→2CoO (Co favors Co₃O₄) ⇒ direct nitridation 2Al+N₂→2AlN at D#14. All replacement constants are exact on the canonical 1-dp table; A#19/B#20 use only enrolled a8 atomic masses.
3. B#7 INSTANCE NOTE: the seed-derived 2N₂O→2N₂+O₂ ratio item was replaced with C₃H₈+5O₂→3CO₂+4H₂O asked O₂:C₃H₈ (5:1) because in the N₂O equation subscript-based reasoning reproduces both the reversal (1:2) and wrong-partner (2:2) values, so no wrong-answer feedback could uniquely name the student's error. In the propane instance every anticipated wrong (1:5 / 5:3 / 1:1) has exactly one producing mistake.
4. TYPED-RATIO LENIENCY (recorded, deliberate): the text-entry ratio contracts accept 'a : b' and 'a to b' alongside the requested colon form — the measured skill is the coefficient pair in the asked order, not the delimiter. sol left this as a minor note on C#2/C#7; consistent bank-wide (B#7 too).
5. C#14 DIAGNOSTICS: the wrong-value diagnostics 0.21 and 0.85 are the exact 2-dp error values; a student answering at the requested 1 dp types 0.2/0.9, which still lands inside the 0.05 matching tolerance. Labels and feedback verified correct — left as-is, flag if your matcher uses exact string compare.
6. GATE_CONFIG ADJUDICATIONS (unchanged through QC, full rationale inside gate_config.json + fix_log.md): a8-taught-constants family (42 enrolled 1-dp masses that recur by the STO parenthetical convention; every occurrence context-verified C1/C2/C3 and honesty-checked against the canonical table), equation-notation gram class (held by the bank-wide identical-equation guard — no two slots share a full equation), spec-verified task idioms + the ratio-colon convention phrase.
7. V-T12 TYPED-WORK CONVENTION (stage-3 ruling, carried): FRQ model answers and rubrics reference substances by worded names where students must type them — nothing typed requires sub/superscripts.
8. FRQ COMPOSITION DEVIATION (stage-1 ruling, flagged then, restating here): frq_per_form=3 OVERRIDES the seed file's frq_per_form=1 baseline per QB-v2 §4 (stoichiometry = calculation-heavy, three distinct taught procedure families: mole-ratio both roles / partial chains / full mass-to-mass chain + inspection). 23 items/form total.
The coefficients of a balanced equation give the reacting ratio for some quantities but not others. For which quantity does the coefficient ratio NOT hold?
Grams of each substance. KEY
Moles of each substance.
Molecules of each substance.
Dozens of molecules of each substance.
STO-001 · easy · recall · seq 2 · seed:U8-G1/pool[0] = STO-001 gateItems[0] (VERBATIM KEEP — C₂H₆ family HOME at seq 2; seed misconceptions carried as anticipatedWrong) · text
In the balanced equation 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O, what is the mole ratio of O₂ to H₂O? Write the ratio with a colon, such as 4:1.
Answer: 7:6 KEY
also accepted: 7 : 6, 7 to 6
wrongValueDiagnostics
[
{
"value": "6:7",
"misconception": "you reversed the order — the question asks O₂ to H₂O, so O₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order. The question asks O₂ to H₂O, so O₂'s coefficient 7 comes first: 7:6."
},
{
"value": "7:4",
"misconception": "you took CO₂'s coefficient instead of H₂O's — the second number must belong to the substance named second",
"feedback": "This answer is wrong — you took CO₂'s coefficient 4 instead of H₂O's. The second number must belong to the substance named second, H₂O, whose coefficient is 6: 7:6."
},
{
"value": "2:6",
"misconception": "you paired C₂H₆'s coefficient with H₂O — neither substance in the asked pair has the coefficient 2",
"feedback": "This answer is wrong — you paired C₂H₆'s coefficient 2 with H₂O. Neither substance in the asked pair has the coefficient 2; O₂'s is 7 and H₂O's is 6: 7:6."
}
]
feedback
{
"keyExplanation": "The coefficient of O₂ is 7, and the coefficient of H₂O is 6. Take the coefficients in the order asked: O₂ first, H₂O second. The mole ratio of O₂ to H₂O is 7:6.",
"anticipatedWrong": [
{
"value": "6:7",
"misconception": "you reversed the order — the question asks O₂ to H₂O, so O₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order. The question asks O₂ to H₂O, so O₂'s coefficient 7 comes first: 7:6."
},
{
"value": "7:4",
"misconception": "you took CO₂'s coefficient instead of H₂O's — the second number must belong to the substance named second",
"feedback": "This answer is wrong — you took CO₂'s coefficient 4 instead of H₂O's. The second number must belong to the substance named second, H₂O, whose coefficient is 6: 7:6."
},
{
"value": "2:6",
"misconception": "you paired C₂H₆'s coefficient with H₂O — neither substance in the asked pair has the coefficient 2",
"feedback": "This answer is wrong — you paired C₂H₆'s coefficient 2 with H₂O. Neither substance in the asked pair has the coefficient 2; O₂'s is 7 and H₂O's is 6: 7:6."
}
]
}
A mass-to-mass calculation converts the given mass into which quantity before the mole ratio can be applied?
Moles. KEY
Grams.
Molar mass.
Coefficients.
STO-004 · medium · application · seq 5 · seed:U8-G1.pool[10] (STO-004 gateItems[1], C₂H₅OH inverted-fraction critique) — instance freshened per must_freshen_instance (ethanol corpus-dead: M=46.0 is a corpus number, family seeds 5 lessons' gates); fuel swapped to C₅H₁₂, critique structure and misconception set kept · mcq4
For C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O, a student writes: moles of H₂O = moles of C₅H₁₂ × (1 ÷ 6). What is wrong with the setup?
The fraction is upside down — the wanted substance H₂O's coefficient 6 belongs on top, giving moles of H₂O = moles of C₅H₁₂ × (6 ÷ 1). KEY
Nothing is wrong — the setup correctly puts the given substance's coefficient on top, so the calculation can go ahead as written.
The only error is the choice of coefficient — CO₂'s 5 should replace the 6, because CO₂ comes first among the products.
The coefficients should be added, giving moles of H₂O = moles of C₅H₁₂ + (6 − 1).
STO-005 · medium · application · seq 6 · seed:U8-G1.pool[13] (STO-005 gateItems[1], blast-furnace Fe₂O₃+3CO) — instance KEPT (equation corpus-free, homed at seq 6 form A); given moles freshened 4.5 → 6.0 per must_freshen_instance (4.5 is a corpus number, STO-005 own text), answer 3.0 → 4.0 · numeric
In a blast furnace, Fe₂O₃ + 3CO → 2Fe + 3CO₂. 6.0 mol of CO reacts completely. How many moles of Fe form? Give your answer in moles to one decimal place.
Answer: 4.0 mol KEY
also accepted: 4.0 mol
wrongValueDiagnostics
[
{
"value": 9.0,
"misconception": "you turned the fraction upside down — the wanted substance Fe's coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Fe's coefficient 2 sits on top."
},
{
"value": 6.0,
"misconception": "you ignored the coefficients and matched the moles one-for-one — the ratio of Fe to CO is 2:3",
"feedback": "This answer is wrong — you ignored the coefficients and matched the moles one-for-one — the ratio of Fe to CO is 2:3."
},
{
"value": 12.0,
"misconception": "you multiplied by Fe's coefficient alone instead of the coefficient ratio",
"feedback": "This answer is wrong — you multiplied by Fe's coefficient alone instead of the coefficient ratio."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of CO = 6.0 mol\nWrite down the equation:\nmoles of Fe = moles of CO × (coefficient of Fe ÷ coefficient of CO)\nSubstitute in the values, and calculate:\nmoles of Fe = 6.0 × (2 ÷ 3)\nmoles of Fe = 4.0 mol"
}
STO-001 · medium · application · seq 7 · seed:U8-G1.pool[1] (STO-001 gateItems[1], Fe₂O₃:Fe 1:2 unwritten-1 + inside-subscript shape) — instance freshened per must_freshen_instance (Fe₂O₃+3CO homed at seq 6) to the SnO₂ + 2H₂ reduction, same 1:2 shape and misconception space · text
In the balanced equation SnO₂ + 2H₂ → Sn + 2H₂O, what is the mole ratio of SnO₂ to H₂? Write the ratio with a colon, such as 4:1.
Answer: 1:2 KEY
also accepted: 1 : 2, 1 to 2
wrongValueDiagnostics
[
{
"value": "2:2",
"misconception": "you used the subscript 2 inside SnO₂ as its coefficient — SnO₂ has no written coefficient, so its coefficient is 1",
"feedback": "This answer is wrong — you used the subscript 2 inside SnO₂ as its coefficient — SnO₂ has no written coefficient, so its coefficient is 1."
},
{
"value": "2:1",
"misconception": "you reversed the order — the question asks SnO₂ to H₂, so SnO₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order — the question asks SnO₂ to H₂, so SnO₂'s coefficient comes first."
},
{
"value": "1:1",
"misconception": "you took Sn's coefficient for H₂ — the second number must belong to H₂, whose coefficient is 2",
"feedback": "This answer is wrong — you took Sn's coefficient for H₂ — the second number must belong to H₂, whose coefficient is 2."
}
]
feedback
{
"keyExplanation": "SnO₂ has no written coefficient, so its coefficient is 1; the coefficient of H₂ is 2. The subscript 2 inside SnO₂ counts atoms in one formula unit and plays no part in the mole ratio. The mole ratio of SnO₂ to H₂ is 1:2.",
"anticipatedWrong": [
{
"value": "2:2",
"misconception": "you used the subscript 2 inside SnO₂ as its coefficient — SnO₂ has no written coefficient, so its coefficient is 1",
"feedback": "This answer is wrong — you used the subscript 2 inside SnO₂ as its coefficient — SnO₂ has no written coefficient, so its coefficient is 1."
},
{
"value": "2:1",
"misconception": "you reversed the order — the question asks SnO₂ to H₂, so SnO₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order — the question asks SnO₂ to H₂, so SnO₂'s coefficient comes first."
},
{
"value": "1:1",
"misconception": "you took Sn's coefficient for H₂ — the second number must belong to H₂, whose coefficient is 2",
"feedback": "This answer is wrong — you took Sn's coefficient for H₂ — the second number must belong to H₂, whose coefficient is 2."
}
]
}
STO-007 · medium · application · seq 8 · seed:U8-G1.pool[18] (STO-007 gateItems[0], 2KClO₃ decomposition shape) — instance freshened per must_freshen_instance + freshness_scope_article (the 2KClO₃ → 2KCl + 3O₂ line is STO-003 fluency corpus verbatim) to the 2Ag₂O → 4Ag + O₂ decomposition; breaks-down-completely framing and wrong space kept · numeric
In 2Ag₂O → 4Ag + O₂, 115.9 g of Ag₂O (M = 231.8 g/mol) breaks down completely. How many moles of O₂ are released? Give your answer in moles to 2 significant figures.
Answer: 0.25 mol KEY
also accepted: 0.25 mol
wrongValueDiagnostics
[
{
"value": 1.0,
"misconception": "you turned the fraction upside down — the wanted substance O₂'s coefficient 1 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance O₂'s coefficient 1 sits on top."
},
{
"value": 0.5,
"misconception": "you skipped the mole-ratio step — 0.50 mol is the Ag₂O's moles, not the O₂'s",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 0.50 mol is the Ag₂O's moles, not the O₂'s."
},
{
"value": 57.95,
"misconception": "you applied the mole ratio to the mass instead of converting to moles first",
"feedback": "This answer is wrong — you applied the mole ratio to the mass instead of converting to moles first."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Ag₂O = 115.9 g, M of Ag₂O = 231.8 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Ag₂O = 115.9 / 231.8\nn of Ag₂O = 0.50 mol\nStep 2 — convert to moles of O₂ with the mole ratio:\nmoles of O₂ = moles of Ag₂O × (coefficient of O₂ ÷ coefficient of Ag₂O)\nmoles of O₂ = 0.50 × (1 ÷ 2)\nmoles of O₂ = 0.25 mol"
}
STO-008 · medium · application · seq 9 · seed:U8-G1.pool[21] (STO-008 gateItems[0], 2KClO3 decomposition) — shape kept (breaks-down decomposition, integer given moles, only wanted M supplied, wrong-M names withheld value), instance freshened to NaHCO3 (2KClO3 equation + M(O2)=32.0 corpus-dead) · numeric
In 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂, 4.0 mol of NaHCO₃ breaks down completely when heated. What mass of Na₂CO₃ forms? (M of Na₂CO₃ = 106.0 g/mol) Give your answer in grams to one decimal place.
Answer: 212.0 g KEY
also accepted: 212.0 g
wrongValueDiagnostics
[
{
"value": 424.0,
"misconception": "you skipped the mole-ratio step — 4.0 mol is the NaHCO₃'s moles, and only half as many moles of Na₂CO₃ form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 4.0 mol is the NaHCO₃'s moles, and only half as many moles of Na₂CO₃ form."
},
{
"value": 848.0,
"misconception": "you turned the fraction upside down — the wanted substance Na₂CO₃'s coefficient 1 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Na₂CO₃'s coefficient 1 sits on top."
},
{
"value": 168.0,
"misconception": "you used NaHCO₃'s molar mass (84.0 g/mol) — step 2 needs the WANTED substance Na₂CO₃'s molar mass",
"feedback": "This answer is wrong — you used NaHCO₃'s molar mass (84.0 g/mol) — step 2 needs the WANTED substance Na₂CO₃'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of NaHCO₃ = 4.0 mol, M of Na₂CO₃ = 106.0 g/mol\nStep 1 — convert to moles of Na₂CO₃ with the mole ratio:\nmoles of Na₂CO₃ = moles of NaHCO₃ × (coefficient of Na₂CO₃ ÷ coefficient of NaHCO₃)\nmoles of Na₂CO₃ = 4.0 × (1 ÷ 2)\nmoles of Na₂CO₃ = 2.0 mol\nStep 2 — convert the moles to mass:\nm = n × M\nm of Na₂CO₃ = 2.0 × 106.0\nm of Na₂CO₃ = 212.0 g"
}
STO-012 · medium · application · seq 10 · seed:U8-G1.pool[33] (STO-012 gateItems[0], zinc-ash defense) KEPT with the two placement-mandated repairs: sample masses re-sized 65.4/81.4/16.0 -> 32.7/40.7/8.0 (16.0 is corpus) and 'oxygen' -> 'the O₂ that reacted' in the key (own-lesson material word) · mcq4
A student watches 32.7 g of zinc burn completely to form 40.7 g of zinc oxide, ZnO, and says the result breaks the law of conservation of mass. Which reply is correct?
It does not — the extra 8.0 g is the O₂ that reacted, and counting both reactants the mass balances exactly. KEY
It does — the law of conservation of mass only holds for reactions run inside sealed containers.
It does not — the law allows small gains when a reaction gives out light and heat.
It does not — zinc oxide is simply a denser substance, and denser substances weigh more.
STO-005 · medium · application · seq 11 · seed:U8-G1.pool[14] (STO-005 gateItems[2], 2C2H6 O2-needed 7:2) — reverse-role O2-needed shape and awkward-ratio class kept, instance freshened to a fresh C8H18 burn (C2H6 homed at seq 2) · numeric
In 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O, how many moles of O₂ are needed to burn 2.0 mol of C₈H₁₈ completely? Give your answer in moles to one decimal place.
Answer: 25.0 mol KEY
also accepted: 25.0 mol
wrongValueDiagnostics
[
{
"value": 0.16,
"misconception": "you turned the fraction upside down — the wanted substance O₂'s coefficient 25 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance O₂'s coefficient 25 sits on top."
},
{
"value": 50.0,
"misconception": "you forgot C₈H₁₈'s coefficient 2 on the bottom of the fraction",
"feedback": "This answer is wrong — you forgot C₈H₁₈'s coefficient 2 on the bottom of the fraction."
},
{
"value": 2.0,
"misconception": "you ignored the coefficients and matched the moles one-for-one — the ratio of O₂ to C₈H₁₈ is 25:2",
"feedback": "This answer is wrong — you ignored the coefficients and matched the moles one-for-one — the ratio of O₂ to C₈H₁₈ is 25:2."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of C₈H₁₈ = 2.0 mol\nWrite down the equation:\nmoles of O₂ = moles of C₈H₁₈ × (coefficient of O₂ ÷ coefficient of C₈H₁₈)\nSubstitute in the values, and calculate:\nmoles of O₂ = 2.0 × (25 ÷ 2)\nmoles of O₂ = 25.0 mol"
}
STO-009 · medium · application · seq 12 · seed:U8-G1.demo_curated[0]/pool[24] (JAMES-CURATED ethanol capstone) — structure, 3-step scaffold shape and misconception set kept verbatim; instance MUST-FRESHEN to a corpus-free metal-halide chain (M(C2H5OH)=46.0 and M(CO2)=44.0 are both corpus numbers — recorded for David/James per the SEED RULING); re-instanced from 2Cr+3S->Cr2S3 after the parallel agent homed that equation at row 20 · numeric
In 2Cr + 3F₂ → 2CrF₃, 456.0 g of F₂ (M = 38.0 g/mol) reacts completely with Cr. What mass of CrF₃ forms? (M of CrF₃ = 109.0 g/mol) Give your answer in grams to one decimal place.
Answer: 872.0 g KEY
also accepted: 872.0 g
wrongValueDiagnostics
[
{
"value": 1308.0,
"misconception": "you skipped the mole-ratio step — 12.0 mol is the F₂'s moles, not the CrF₃'s",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 12.0 mol is the F₂'s moles, not the CrF₃'s."
},
{
"value": 1962.0,
"misconception": "you turned the fraction upside down — the wanted substance CrF₃'s coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance CrF₃'s coefficient 2 sits on top."
},
{
"value": 304.0,
"misconception": "you used F₂'s molar mass in step 3 — the final step needs the WANTED substance CrF₃'s molar mass",
"feedback": "This answer is wrong — you used F₂'s molar mass in step 3 — the final step needs the WANTED substance CrF₃'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of F₂ = 456.0 g, M of F₂ = 38.0 g/mol, M of CrF₃ = 109.0 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of F₂ = 456.0 / 38.0\nn of F₂ = 12.0 mol\nStep 2 — convert to moles of CrF₃ with the mole ratio:\nmoles of CrF₃ = moles of F₂ × (coefficient of CrF₃ ÷ coefficient of F₂)\nmoles of CrF₃ = 12.0 × (2 ÷ 3)\nmoles of CrF₃ = 8.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of CrF₃ = 8.0 × 109.0\nm of CrF₃ = 872.0 g"
}
STO-011 · medium · application · seq 13 · seed:U8-G1 STO-011 gateItems[1] (4Fe+3O2 total-products mcq) — CONVERTED to numeric ENTRY per the HS numeric-ENTRY rule, distractors carried over as anticipatedWrong misconceptions; instance freshened (4Fe+3O2->2Fe2O3 is STO-012 fluency corpus; 32.0/159.6 corpus-banned; fresh instance 4V + 5O2 -> 2V2O5) · numeric
Vanadium burns in O₂: 4V + 5O₂ → 2V₂O₅. (M of V₂O₅ = 181.8 g/mol.) What total mass of products does the balanced equation represent? Give your answer in grams to one decimal place.
Answer: 363.6 g KEY
also accepted: 363.6 g
wrongValueDiagnostics
[
{
"value": 181.8,
"misconception": "you ignored the coefficient 2 on V₂O₅ — its mass is 2 × 181.8 g",
"feedback": "This answer is wrong — you ignored the coefficient 2 on V₂O₅ — its mass is 2 × 181.8 g."
},
{
"value": 909.0,
"misconception": "you multiplied by 5, the O₂ coefficient, instead of by V₂O₅'s own coefficient 2",
"feedback": "This answer is wrong — you multiplied by 5, the O₂ coefficient, instead of by V₂O₅'s own coefficient 2."
},
{
"value": 183.8,
"misconception": "you added the coefficient instead of multiplying by it",
"feedback": "This answer is wrong — you added the coefficient instead of multiplying by it."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM of V₂O₅ = 181.8 g/mol, coefficient of V₂O₅ = 2\nWrite down the equation:\nmass of each substance = coefficient × molar mass\nSubstitute in the values, and calculate:\ntotal mass of products = 2 × 181.8\ntotal mass of products = 363.6 g"
}
STO-007 · medium · application · seq 14 · seed:U8-G1 STO-007 gateItems[2] (N2+O2->2NO shape) — instance freshened (equation is STO-002 WE0 corpus; 28.0 corpus-banned) and rotated to the PRODUCT-mass-given read per the blueprint case rotation (fresh instance 4Cu + O2 -> 2Cu2O; Cu spoken as SYMBOL, copper word homed seq 10-B) · numeric
Cu is heated in O₂ and reacts completely: 4Cu + O₂ → 2Cu₂O. The reaction forms 443.3 g of Cu₂O (M = 143.0 g/mol). How many moles of Cu reacted? Give your answer in moles to one decimal place.
Answer: 6.2 mol KEY
also accepted: 6.2 mol
wrongValueDiagnostics
[
{
"value": 1.55,
"misconception": "you turned the fraction upside down — the wanted substance Cu's coefficient 4 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Cu's coefficient 4 sits on top."
},
{
"value": 3.1,
"misconception": "you stopped at step 1 — 3.1 mol is the Cu₂O's moles, not the Cu's",
"feedback": "This answer is wrong — you stopped at step 1 — 3.1 mol is the Cu₂O's moles, not the Cu's."
},
{
"value": 886.6,
"misconception": "you applied the mole ratio to the mass instead of converting to moles first",
"feedback": "This answer is wrong — you applied the mole ratio to the mass instead of converting to moles first."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Cu₂O = 443.3 g, M of Cu₂O = 143.0 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Cu₂O = 443.3 / 143.0\nn of Cu₂O = 3.1 mol\nStep 2 — convert to moles of Cu with the mole ratio:\nmoles of Cu = moles of Cu₂O × (coefficient of Cu ÷ coefficient of Cu₂O)\nmoles of Cu = 3.1 × (4 ÷ 2)\nmoles of Cu = 6.2 mol"
}
STO-010 · medium · application · seq 15 · seed:U8-G1 STO-010 gateItems[1] (H2O2 demonstration-target) — instance freshened (H2O2 is STO-007 WE0 + STO-013 WE0 corpus; 34.0/32.0 corpus-banned); target framing kept on the fresh instance 2Sb + 3F2 -> 2SbF3 (F2-partner precedent from the authored bank) · numeric
A laboratory preparation needs 447.0 g of SbF₃. In 2Sb + 3F₂ → 2SbF₃, what mass of F₂ must react completely with antimony to supply it? (M of SbF₃ = 178.8 g/mol; M of F₂ = 38.0 g/mol) Give your answer in grams to one decimal place.
Answer: 142.5 g KEY
also accepted: 142.5 g
wrongValueDiagnostics
[
{
"value": 95.0,
"misconception": "you skipped the mole-ratio step — 2.5 mol is the SbF₃'s moles, and more moles of F₂ are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 2.5 mol is the SbF₃'s moles, and more moles of F₂ are needed."
},
{
"value": 63.3,
"misconception": "you turned the fraction upside down — the wanted substance F₂'s coefficient 3 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance F₂'s coefficient 3 sits on top."
},
{
"value": 670.5,
"misconception": "you used SbF₃'s molar mass in step 3 — the final step needs the WANTED substance F₂'s molar mass",
"feedback": "This answer is wrong — you used SbF₃'s molar mass in step 3 — the final step needs the WANTED substance F₂'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of SbF₃ = 447.0 g, M of SbF₃ = 178.8 g/mol, M of F₂ = 38.0 g/mol\nStep 1 — convert the target mass to moles:\nn = m / M\nn of SbF₃ = 447.0 / 178.8\nn of SbF₃ = 2.5 mol\nStep 2 — convert to moles of F₂ with the mole ratio:\nmoles of F₂ = moles of SbF₃ × (coefficient of F₂ ÷ coefficient of SbF₃)\nmoles of F₂ = 2.5 × (3 ÷ 2)\nmoles of F₂ = 3.75 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of F₂ = 3.75 × 38.0\nm of F₂ = 142.5 g"
}
STO-008 · medium · application · seq 16 · fully authored (lesson F3 reverse pairing per the blueprint: mass of the OTHER substance for an integer target moles; wanted-M-only supplied, wrong-M misconception names the withheld M(PCl3)=137.5; fresh instance P4 + 6Cl2 -> 4PCl3) · numeric
In P₄ + 6Cl₂ → 4PCl₃, what mass of P₄ is needed to make 2.0 mol of PCl₃? (M of P₄ = 124.0 g/mol) Give your answer in grams to one decimal place.
Answer: 62.0 g KEY
also accepted: 62.0 g
wrongValueDiagnostics
[
{
"value": 248.0,
"misconception": "you skipped the mole-ratio step — 2.0 mol is the PCl₃'s moles, and only a quarter as many moles of P₄ are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 2.0 mol is the PCl₃'s moles, and only a quarter as many moles of P₄ are needed."
},
{
"value": 992.0,
"misconception": "you turned the fraction upside down — the wanted substance P₄'s coefficient 1 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance P₄'s coefficient 1 sits on top."
},
{
"value": 68.8,
"misconception": "you used PCl₃'s molar mass (137.5 g/mol) — step 2 needs the WANTED substance P₄'s molar mass",
"feedback": "This answer is wrong — you used PCl₃'s molar mass (137.5 g/mol) — step 2 needs the WANTED substance P₄'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of PCl₃ = 2.0 mol, M of P₄ = 124.0 g/mol\nStep 1 — convert to moles of P₄ with the mole ratio:\nmoles of P₄ = moles of PCl₃ × (coefficient of P₄ ÷ coefficient of PCl₃)\nmoles of P₄ = 2.0 × (1 ÷ 4)\nmoles of P₄ = 0.50 mol\nStep 2 — convert the moles to mass:\nm = n × M\nm of P₄ = 0.50 × 124.0\nm of P₄ = 62.0 g"
}
STO-011 · hard · transfer · seq 17 · seed:U8-G1.pool[32] (STO-011 gateItems[2], C2H4 audit) — shape and distractor classes kept, instance freshened per must_freshen_instance (M parenthetical carried corpus numbers 28.0/32.0/44.0/18.0); error class: product coefficients ignored · mcq4
For 3Zn + 2CrCl₃ → 3ZnCl₂ + 2Cr, a student totals the two sides and gets 513.2 g of reactants but only 188.4 g of products. (M of Zn = 65.4 g/mol, M of CrCl₃ = 158.5 g/mol, M of ZnCl₂ = 136.4 g/mol, M of Cr = 52.0 g/mol.) What went wrong?
The product coefficients were ignored — the product side is 3 × 136.4 + 2 × 52.0 = 513.2 g. KEY
The reactant total is wrong — the Zn needs a coefficient greater than 3 to balance the chlorine.
The equation itself is unbalanced, so the totals cannot match.
Nothing went wrong — a displacement reaction always loses mass as the metals swap places.
STO-013 · hard · transfer · seq 18 · seed:U8-G1.pool[36] (STO-013 gateItems[0], CaCO3 step-3 wrong-M) — class and option format kept, instance freshened per must_freshen_instance (CaCO3 is STO-011 WE1 corpus; 100.1/44.0 corpus numbers) · mcq4
For Sn + O₂ → SnO₂, a student calculates the mass of SnO₂ formed from 356.1 g of Sn. (M of Sn = 118.7 g/mol, M of SnO₂ = 150.7 g/mol.)
Step 1: n of Sn = 356.1 / 118.7 = 3.0 mol
Step 2: moles of SnO₂ = 3.0 × 1/1 = 3.0 mol
Step 3: m of SnO₂ = 3.0 × 118.7 = 356.1 g
Which step contains the error?
Step 3 KEY
Step 1
Step 2
No step contains an error
STO-010 · hard · transfer · seq 19 · fully authored per blueprint (reverse chain product-target -> reactant mass, awkward 2:1 pair, both Ms supplied — both enrolled a8 atomic constants; wrongs: skipped ratio / inverted / wrong-M; wave-10 sol realism fix: 4Na+O2->2Na2O dropped (Na combustion favors the peroxide; glassmaking sources Na2O from soda ash) => Kroll refining step TiCl4 + 2Mg -> Ti + 2MgCl2; airbag-NaN3 and n=6 repicks census-killed) · numeric
In the refining step TiCl₄ + 2Mg → Ti + 2MgCl₂, a metals plant must produce 383.2 g of Ti in one batch. What mass of Mg must react completely to make it? (M of Ti = 47.9 g/mol; M of Mg = 24.3 g/mol) Give your answer in grams to one decimal place.
Answer: 388.8 g KEY
also accepted: 388.8 g
wrongValueDiagnostics
[
{
"value": 194.4,
"misconception": "you skipped the mole-ratio step — 8.0 mol is the Ti's moles, and twice as many moles of Mg are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 8.0 mol is the Ti's moles, and twice as many moles of Mg are needed."
},
{
"value": 97.2,
"misconception": "you turned the fraction upside down — the wanted substance Mg's coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Mg's coefficient 2 sits on top."
},
{
"value": 766.4,
"misconception": "you used Ti's molar mass in step 3 — the final step needs the WANTED substance Mg's molar mass",
"feedback": "This answer is wrong — you used Ti's molar mass in step 3 — the final step needs the WANTED substance Mg's molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Ti = 383.2 g, M of Ti = 47.9 g/mol, M of Mg = 24.3 g/mol\nStep 1 — convert the target mass to moles:\nn = m / M\nn of Ti = 383.2 / 47.9\nn of Ti = 8.0 mol\nStep 2 — convert to moles of Mg with the mole ratio:\nmoles of Mg = moles of Ti × (coefficient of Mg ÷ coefficient of Ti)\nmoles of Mg = 8.0 × (2 ÷ 1)\nmoles of Mg = 16.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of Mg = 16.0 × 24.3\nm of Mg = 388.8 g"
}
STO-009 · hard · transfer · seq 20 · fully authored per blueprint (capstone terminal slot: full forward chain, awkward 3:1 pair, transfer ceramic surface; reactant→product only per the STO-010 boundary) · numeric
Si₃N₄ is a hard ceramic used for engine bearings. In 3Si + 2N₂ → Si₃N₄, 168.6 g of Si (M = 28.1 g/mol) reacts completely with N₂. What mass of Si₃N₄ forms? (M of Si₃N₄ = 140.3 g/mol) Give your answer in grams to one decimal place.
Answer: 280.6 g KEY
also accepted: 280.6 g
wrongValueDiagnostics
[
{
"value": 841.8,
"misconception": "you skipped the mole-ratio step — 6.0 mol is the Si's moles, and only a third as many moles of Si₃N₄ form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 6.0 mol is the Si's moles, and only a third as many moles of Si₃N₄ form."
},
{
"value": 2525.4,
"misconception": "you turned the fraction upside down — the wanted substance Si₃N₄'s coefficient 1 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Si₃N₄'s coefficient 1 sits on top."
},
{
"value": 56.2,
"misconception": "you used Si's molar mass in step 3 — the final step needs the WANTED substance Si₃N₄'s molar mass",
"feedback": "This answer is wrong — you used Si's molar mass in step 3 — the final step needs the WANTED substance Si₃N₄'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Si = 168.6 g, M of Si = 28.1 g/mol, M of Si₃N₄ = 140.3 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Si = 168.6 / 28.1\nn of Si = 6.0 mol\nStep 2 — convert to moles of Si₃N₄ with the mole ratio:\nmoles of Si₃N₄ = moles of Si × (coefficient of Si₃N₄ ÷ coefficient of Si)\nmoles of Si₃N₄ = 6.0 × (1 ÷ 3)\nmoles of Si₃N₄ = 2.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of Si₃N₄ = 2.0 × 140.3\nm of Si₃N₄ = 280.6 g"
}
STO-005 · medium · application · seq 21 · authored (FRQ1 per composition.frq_split_rationale: STO-005 mole-to-mole in BOTH roles + taught sanity check; mole-only, no molar masses per the STO-007+ boundary; fresh reaction + numbers per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
Al reacts with Cl₂ according to the balanced equation 2Al + 3Cl₂ → 2AlCl₃. Work in moles throughout — no molar masses are needed.
Al reacts with Cl₂ according to the balanced equation 2Al + 3Cl₂ → 2AlCl₃. Work in moles throughout — no molar masses are needed.
(a) 6.0 mol of Cl₂ reacts completely with Al. Calculate the moles of AlCl₃ that form. Show your working: state the given moles, write the mole-ratio equation with both substances named, substitute, and calculate.
(b) Calculate the moles of Cl₂ needed to make 8.0 mol of AlCl₃. Show the same working.
(c) For part (a), apply the lesson's sanity check without recalculating: compare the coefficient of the wanted substance with the coefficient of the given substance, state whether the answer should come out smaller or larger than the given moles, and confirm that your part (a) answer passes the check.
Part a. 6.0 mol of Cl₂ reacts completely with Al. Calculate the moles of AlCl₃ that form. Show your working: state the given moles, write the mole-ratio equation with both substances named, substitute, and calculate.
Model: moles of chlorine = 6.0 mol
moles of AlCl3 = moles of chlorine × (coefficient of AlCl3 ÷ coefficient of chlorine)
moles of AlCl3 = 6.0 × (2 ÷ 3)
moles of AlCl3 = 4.0 mol
Part b. Calculate the moles of Cl₂ needed to make 8.0 mol of AlCl₃. Show the same working.
Model: moles of AlCl3 = 8.0 mol (the target)
moles of chlorine = moles of AlCl3 × (coefficient of chlorine ÷ coefficient of AlCl3)
moles of chlorine = 8.0 × (3 ÷ 2)
moles of chlorine = 12.0 mol
Part c. For part (a), apply the lesson's sanity check without recalculating: compare the coefficient of the wanted substance with the coefficient of the given substance, state whether the answer should come out smaller or larger than the given moles, and confirm that your part (a) answer passes the check.
Model: The taught check says: when the wanted coefficient is smaller than the given one, the answer comes out smaller than the given moles. In part (a) the wanted substance AlCl3 has coefficient 2 and the given substance chlorine has coefficient 3 — the wanted coefficient is smaller, so the answer should come out smaller than 6.0 mol. The answer 4.0 mol is smaller than 6.0 mol, so it passes the check.
Full model answer. (a)
moles of chlorine = 6.0 mol
moles of AlCl3 = moles of chlorine × (coefficient of AlCl3 ÷ coefficient of chlorine)
moles of AlCl3 = 6.0 × (2 ÷ 3)
moles of AlCl3 = 4.0 mol
(b)
moles of AlCl3 = 8.0 mol (the target)
moles of chlorine = moles of AlCl3 × (coefficient of chlorine ÷ coefficient of AlCl3)
moles of chlorine = 8.0 × (3 ÷ 2)
moles of chlorine = 12.0 mol
(c)
The taught check says: when the wanted coefficient is smaller than the given one, the answer comes out smaller than the given moles. In part (a) the wanted substance AlCl3 has coefficient 2 and the given substance chlorine has coefficient 3 — the wanted coefficient is smaller, so the answer should come out smaller than 6.0 mol. The answer 4.0 mol is smaller than 6.0 mol, so it passes the check.
STO-009 · hard · application · seq 22 · authored (FRQ2 per composition.frq_split_rationale: STO-009 capstone chain end-to-end — plan + full mass-to-mass run + step-3 molar-mass rule; forward reactant→product only per the STO-010 boundary; both Ms supplied, corpus-free; fresh reaction + masses per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
Fe burns in pure O₂ to form Fe₃O₄ as the balanced equation shows 3Fe + 2O₂ → Fe₃O₄. 334.8 g of Fe reacts completely. (M of Fe = 55.8 g/mol; M of Fe₃O₄ = 231.4 g/mol.)
Fe burns in pure O₂ to form Fe₃O₄ as the balanced equation shows 3Fe + 2O₂ → Fe₃O₄. 334.8 g of Fe reacts completely. (M of Fe = 55.8 g/mol; M of Fe₃O₄ = 231.4 g/mol.)
(a) Before calculating, state the three-step plan for working out the mass of Fe₃O₄ that forms, naming the tool each step uses.
(b) Carry out the calculation, showing the working for every step, and give the mass of Fe₃O₄ that forms.
(c) State which substance's molar mass belongs in step 3, and why.
Part a. Before calculating, state the three-step plan for working out the mass of Fe₃O₄ that forms, naming the tool each step uses.
Model: grams of Fe → moles of Fe (n = m/M) → moles of the iron oxide (mole-ratio equation) → grams of the iron oxide (m = n × M).
Part b. Carry out the calculation, showing the working for every step, and give the mass of Fe₃O₄ that forms.
Model: m of Fe = 334.8 g; M of Fe = 55.8 g/mol; M of the iron oxide = 231.4 g/mol
Step 1 — convert the mass to moles:
n = m / M
n of Fe = 334.8 / 55.8
n of Fe = 6.0 mol
Step 2 — convert to moles of the iron oxide with the mole ratio:
moles of the iron oxide = moles of Fe × (coefficient of the iron oxide ÷ coefficient of Fe)
moles of the iron oxide = 6.0 × (1 ÷ 3)
moles of the iron oxide = 2.0 mol
Step 3 — convert the moles to mass:
m = n × M
m of the iron oxide = 2.0 × 231.4
m of the iron oxide = 462.8 g
Part c. State which substance's molar mass belongs in step 3, and why.
Model: Step 3 must use the WANTED substance's molar mass — the iron oxide's, 231.4 g/mol — because step 3 converts the wanted substance's moles into its mass with m = n × M. The given substance's molar mass (Fe's 55.8 g/mol) is used only in step 1 and never appears in step 3.
Full model answer. (a)
grams of Fe → moles of Fe (n = m/M) → moles of the iron oxide (mole-ratio equation) → grams of the iron oxide (m = n × M).
(b)
m of Fe = 334.8 g; M of Fe = 55.8 g/mol; M of the iron oxide = 231.4 g/mol
Step 1 — convert the mass to moles:
n = m / M
n of Fe = 334.8 / 55.8
n of Fe = 6.0 mol
Step 2 — convert to moles of the iron oxide with the mole ratio:
moles of the iron oxide = moles of Fe × (coefficient of the iron oxide ÷ coefficient of Fe)
moles of the iron oxide = 6.0 × (1 ÷ 3)
moles of the iron oxide = 2.0 mol
Step 3 — convert the moles to mass:
m = n × M
m of the iron oxide = 2.0 × 231.4
m of the iron oxide = 462.8 g
(c)
Step 3 must use the WANTED substance's molar mass — the iron oxide's, 231.4 g/mol — because step 3 converts the wanted substance's moles into its mass with m = n × M. The given substance's molar mass (Fe's 55.8 g/mol) is used only in step 1 and never appears in step 3.
STO-013 · hard · transfer · seq 23 · authored (FRQ3 per composition.frq_split_rationale: STO-013 inspection PRODUCTION — one planted error; A-form class = step-2 flipped ratio (the taught most-common error); fresh reaction + numbers per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
For Fe₃O₄ + 4H₂ → 3Fe + 4H₂O, a student calculates the mass of Fe formed when 694.2 g of Fe₃O₄ reacts completely. (M of Fe₃O₄ = 231.4 g/mol; M of Fe = 55.8 g/mol.)
Step 1: n of Fe₃O₄ = 694.2 / 231.4 = 3.0 mol
Step 2: moles of Fe = 3.0 × (1 ÷ 3) = 1.0 mol
Step 3: m of Fe = 1.0 × 55.8 = 55.8 g
Exactly one step contains an error.
For Fe₃O₄ + 4H₂ → 3Fe + 4H₂O, a student calculates the mass of Fe formed when 694.2 g of Fe₃O₄ reacts completely. (M of Fe₃O₄ = 231.4 g/mol; M of Fe = 55.8 g/mol.)
Step 1: n of Fe₃O₄ = 694.2 / 231.4 = 3.0 mol
Step 2: moles of Fe = 3.0 × (1 ÷ 3) = 1.0 mol
Step 3: m of Fe = 1.0 × 55.8 = 55.8 g
Exactly one step contains an error.
(a) Name the step that contains the error, and state what is wrong with it.
(b) Correct the erroneous step, rerun the working from that point, and give the corrected mass of Fe.
(c) A classmate claims step 3 must also contain an error, because it produced a wrong mass. Using the lesson's rule about steps that inherit a wrong number, state whether the classmate is right, and explain.
Part a. Name the step that contains the error, and state what is wrong with it.
Model: The error is in step 2. Checking each step: step 1 divides 694.2 g by the iron oxide's own molar mass, 231.4 g/mol — correct. In step 2 the balanced equation requires the mole ratio coefficient of Fe ÷ coefficient of the iron oxide = 3 ÷ 1, but the student wrote 1 ÷ 3 — the ratio is flipped.
Part b. Correct the erroneous step, rerun the working from that point, and give the corrected mass of Fe.
Model: Corrected step 2:
moles of the iron oxide = 3.0 mol
moles of Fe = moles of the iron oxide × (coefficient of Fe ÷ coefficient of the iron oxide)
moles of Fe = 3.0 × (3 ÷ 1)
moles of Fe = 9.0 mol
Rerun step 3:
m = n × M
m of Fe = 9.0 × 55.8
m of Fe = 502.2 g
Part c. A classmate claims step 3 must also contain an error, because it produced a wrong mass. Using the lesson's rule about steps that inherit a wrong number, state whether the classmate is right, and explain.
Model: Step 3 does not contain an error. The taught rule: a step that inherits a wrong number from an earlier step is not itself the error — the error lives where the wrong number was made. Step 3 used the correct method and the correct molar mass (Fe's 55.8 g/mol); it only inherited the wrong 1.0 mol that step 2 made, so the error lives in step 2, not step 3.
Full model answer. (a)
The error is in step 2. Checking each step: step 1 divides 694.2 g by the iron oxide's own molar mass, 231.4 g/mol — correct. In step 2 the balanced equation requires the mole ratio coefficient of Fe ÷ coefficient of the iron oxide = 3 ÷ 1, but the student wrote 1 ÷ 3 — the ratio is flipped.
(b)
Corrected step 2:
moles of the iron oxide = 3.0 mol
moles of Fe = moles of the iron oxide × (coefficient of Fe ÷ coefficient of the iron oxide)
moles of Fe = 3.0 × (3 ÷ 1)
moles of Fe = 9.0 mol
Rerun step 3:
m = n × M
m of Fe = 9.0 × 55.8
m of Fe = 502.2 g
(c)
Step 3 does not contain an error. The taught rule: a step that inherits a wrong number from an earlier step is not itself the error — the error lives where the wrong number was made. Step 3 used the correct method and the correct molar mass (Fe's 55.8 g/mol); it only inherited the wrong 1.0 mol that step 2 made, so the error lives in step 2, not step 3.
The balanced equation 2Na + Cl₂ → 2NaCl can be read in more than one counting unit. Which reading is correct for the correct reason?
2 mol of Na react with 1 mol of Cl₂ — because a mole is the same fixed count for every substance, so scaling every substance up by that same count leaves the ratio unchanged. KEY
2 g of Na react with 1 g of Cl₂ — because the coefficients fix the reacting masses directly, so exactly the same two numbers carry straight over into the gram amounts.
2 mol of Na react with 1 mol of Cl₂ — because Na and Cl₂ have equal molar masses, so equal counts always pair up.
2 mol of Na react with 1 mol of Cl₂ — because the mole is defined separately for each reaction so that its coefficients come out right.
In the balanced equation 2Ni + O₂ → 2NiO, what is the mole ratio of NiO to O₂? Write the ratio with a colon, such as 4:1.
Answer: 2:1 KEY
also accepted: 2 : 1, 2 to 1
wrongValueDiagnostics
[
{
"value": "1:2",
"misconception": "you reversed the order — the question asks NiO to O₂, so NiO's coefficient comes first",
"feedback": "This answer is wrong — you reversed the order. The question asks NiO to O₂, so NiO's coefficient 2 comes first: 2:1."
},
{
"value": "2:2",
"misconception": "you used the subscript 2 inside O₂ as its coefficient — O₂ has no written coefficient, so its coefficient is 1",
"feedback": "This answer is wrong — you used the subscript 2 inside O₂ as its coefficient. Subscripts play no part; O₂ has no written coefficient, so its coefficient is 1: 2:1."
},
{
"value": "1:1",
"misconception": "you ignored the coefficients — the mole ratio is read from the written coefficients, and NiO's is 2",
"feedback": "This answer is wrong — you ignored the coefficients. The mole ratio comes from the coefficients: NiO's is 2 and O₂'s is the unwritten 1, so the ratio is 2:1."
}
]
feedback
{
"keyExplanation": "The coefficient of NiO is 2, and O₂ has no written coefficient, so its coefficient is 1. Take the coefficients in the order asked: NiO first, O₂ second. The mole ratio of NiO to O₂ is 2:1.",
"anticipatedWrong": [
{
"value": "1:2",
"misconception": "you reversed the order — the question asks NiO to O₂, so NiO's coefficient comes first",
"feedback": "This answer is wrong — you reversed the order. The question asks NiO to O₂, so NiO's coefficient 2 comes first: 2:1."
},
{
"value": "2:2",
"misconception": "you used the subscript 2 inside O₂ as its coefficient — O₂ has no written coefficient, so its coefficient is 1",
"feedback": "This answer is wrong — you used the subscript 2 inside O₂ as its coefficient. Subscripts play no part; O₂ has no written coefficient, so its coefficient is 1: 2:1."
},
{
"value": "1:1",
"misconception": "you ignored the coefficients — the mole ratio is read from the written coefficients, and NiO's is 2",
"feedback": "This answer is wrong — you ignored the coefficients. The mole ratio comes from the coefficients: NiO's is 2 and O₂'s is the unwritten 1, so the ratio is 2:1."
}
]
}
A student wants the mole ratio of Mg to O₂ and finds two versions of the equation: the unbalanced Mg + O₂ → MgO and the balanced 2Mg + O₂ → 2MgO. Why must the ratio be read from the balanced version?
Only the balanced coefficients count atoms correctly — the unbalanced version shows atoms disappearing, and a reaction only rearranges atoms. KEY
The unbalanced version works for small amounts — its coefficients only begin to fail once larger amounts of Mg are reacting.
The unbalanced version puts the substances on the wrong sides of the arrow, so its ratio comes out reversed.
The unbalanced version gives exactly half of every true coefficient, so doubling every number would also fix it.
STO-006 · easy · recall · seq 4 · authored (hub-recognition rotation per blueprint, instance-free, zero arithmetic per JM-P1a; bare-quantity options parallel to the seed's format) · mcq4
In a mass-to-mass calculation, the step that crosses from the given substance to the wanted substance works in which quantity?
In 2Sb + 3Cl₂ → 2SbCl₃, 12.0 mol of Cl₂ reacts completely with Sb. How many moles of SbCl₃ form? Give your answer in moles to one decimal place.
Answer: 8.0 mol KEY
also accepted: 8.0 mol
wrongValueDiagnostics
[
{
"value": 18.0,
"misconception": "you turned the fraction upside down — the wanted substance SbCl₃'s coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance SbCl₃'s coefficient 2 sits on top."
},
{
"value": 12.0,
"misconception": "you ignored the coefficients and matched the moles one-for-one — the ratio of SbCl₃ to Cl₂ is 2:3",
"feedback": "This answer is wrong — you ignored the coefficients and matched the moles one-for-one — the ratio of SbCl₃ to Cl₂ is 2:3."
},
{
"value": 24.0,
"misconception": "you multiplied by SbCl₃'s coefficient alone instead of the coefficient ratio",
"feedback": "This answer is wrong — you multiplied by SbCl₃'s coefficient alone instead of the coefficient ratio."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of Cl₂ = 12.0 mol\nWrite down the equation:\nmoles of SbCl₃ = moles of Cl₂ × (coefficient of SbCl₃ ÷ coefficient of Cl₂)\nSubstitute in the values, and calculate:\nmoles of SbCl₃ = 12.0 × (2 ÷ 3)\nmoles of SbCl₃ = 8.0 mol"
}
STO-001 · medium · application · seq 7 · seed:U8-G1.pool[2] (STO-001 gateItems[2], SO₃:O₂ 2:1 unwritten-1 shape) — wave-11 sol fix: the freshened 2N₂O instance had a DEGENERATE wrong-value space (subscript routes reproduce the reversal and wrong-partner values, so feedback over-attributes the error — sol loop-1+2 fails); re-freshened to C₃H₈ + 5O₂ → 3CO₂ + 4H₂O asked O₂:C₃H₈ (unwritten-1 shape kept; wrongs 1:5 / 5:3 / 1:1 each uniquely attributable; key 5:1 unreachable by any subscript reading) · text
In the balanced equation C₃H₈ + 5O₂ → 3CO₂ + 4H₂O, what is the mole ratio of O₂ to C₃H₈? Write the ratio with a colon, such as 4:1.
Answer: 5:1 KEY
also accepted: 5 : 1, 5 to 1
wrongValueDiagnostics
[
{
"value": "1:5",
"misconception": "you reversed the order — the question asks O₂ to C₃H₈, so O₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order — the question asks O₂ to C₃H₈, so O₂'s coefficient comes first."
},
{
"value": "5:3",
"misconception": "you paired O₂ with CO₂'s coefficient — the asked partner is C₃H₈, whose unwritten coefficient is 1",
"feedback": "This answer is wrong — you paired O₂ with CO₂'s coefficient — the asked partner is C₃H₈, whose unwritten coefficient is 1."
},
{
"value": "1:1",
"misconception": "you matched the substances one-for-one and ignored the coefficients — O₂'s coefficient is 5, and C₃H₈'s unwritten coefficient is 1",
"feedback": "This answer is wrong — you matched the substances one-for-one and ignored the coefficients — O₂'s coefficient is 5, and C₃H₈'s unwritten coefficient is 1."
}
]
feedback
{
"keyExplanation": "The coefficient of O₂ is 5, and C₃H₈ has no written coefficient, so its coefficient is 1. The subscripts inside the formulas play no part. Take the coefficients in the order asked: O₂ first, C₃H₈ second. The mole ratio of O₂ to C₃H₈ is 5:1.",
"anticipatedWrong": [
{
"value": "1:5",
"misconception": "you reversed the order — the question asks O₂ to C₃H₈, so O₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order — the question asks O₂ to C₃H₈, so O₂'s coefficient comes first."
},
{
"value": "5:3",
"misconception": "you paired O₂ with CO₂'s coefficient — the asked partner is C₃H₈, whose unwritten coefficient is 1",
"feedback": "This answer is wrong — you paired O₂ with CO₂'s coefficient — the asked partner is C₃H₈, whose unwritten coefficient is 1."
},
{
"value": "1:1",
"misconception": "you matched the substances one-for-one and ignored the coefficients — O₂'s coefficient is 5, and C₃H₈'s unwritten coefficient is 1",
"feedback": "This answer is wrong — you matched the substances one-for-one and ignored the coefficients — O₂'s coefficient is 5, and C₃H₈'s unwritten coefficient is 1."
}
]
}
STO-007 · medium · application · seq 8 · authored (parallel numeric slot; fresh metal + halogen surface per the case rotation; only the given substance's M supplied; wave-5 fresh instance 2Li + Cl₂ → 2LiCl) · numeric
In 2Li + Cl₂ → 2LiCl, 177.5 g of Cl₂ (M = 71.0 g/mol) reacts completely with Li. How many moles of LiCl form? Give your answer in moles to 2 significant figures.
Answer: 5.0 mol KEY
also accepted: 5.0 mol
wrongValueDiagnostics
[
{
"value": 1.25,
"misconception": "you turned the fraction upside down — the wanted substance LiCl's coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance LiCl's coefficient 2 sits on top."
},
{
"value": 2.5,
"misconception": "you skipped the mole-ratio step — 2.5 mol is the Cl₂'s moles, not the LiCl's",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 2.5 mol is the Cl₂'s moles, not the LiCl's."
},
{
"value": 355.0,
"misconception": "you applied the mole ratio to the mass instead of converting to moles first",
"feedback": "This answer is wrong — you applied the mole ratio to the mass instead of converting to moles first."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Cl₂ = 177.5 g, M of Cl₂ = 71.0 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Cl₂ = 177.5 / 71.0\nn of Cl₂ = 2.5 mol\nStep 2 — convert to moles of LiCl with the mole ratio:\nmoles of LiCl = moles of Cl₂ × (coefficient of LiCl ÷ coefficient of Cl₂)\nmoles of LiCl = 2.5 × (2 ÷ 1)\nmoles of LiCl = 5.0 mol"
}
In 2Ba + O₂ → 2BaO, 6.0 mol of O₂ reacts completely with Ba. What mass of BaO forms? (M of BaO = 153.3 g/mol) Give your answer in grams to one decimal place.
Answer: 1839.6 g KEY
also accepted: 1839.6 g, 1,839.6
wrongValueDiagnostics
[
{
"value": 919.8,
"misconception": "you skipped the mole-ratio step — 6.0 mol is the O₂'s moles, and twice as many moles of BaO form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 6.0 mol is the O₂'s moles, and twice as many moles of BaO form."
},
{
"value": 459.9,
"misconception": "you turned the fraction upside down — the wanted substance BaO's coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance BaO's coefficient 2 sits on top."
},
{
"value": 384.0,
"misconception": "you used O₂'s molar mass (32.0 g/mol) — step 2 needs the WANTED substance BaO's molar mass",
"feedback": "This answer is wrong — you used O₂'s molar mass (32.0 g/mol) — step 2 needs the WANTED substance BaO's molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of O₂ = 6.0 mol, M of BaO = 153.3 g/mol\nStep 1 — convert to moles of BaO with the mole ratio:\nmoles of BaO = moles of O₂ × (coefficient of BaO ÷ coefficient of O₂)\nmoles of BaO = 6.0 × (2 ÷ 1)\nmoles of BaO = 12.0 mol\nStep 2 — convert the moles to mass:\nm = n × M\nm of BaO = 12.0 × 153.3\nm of BaO = 1839.6 g"
}
STO-012 · medium · application · seq 10 · seed:U8-G1.pool[34] (STO-012 gateItems[1], copper prediction) KEPT with the one placement-mandated repair: 'the oxygen that reacts' -> 'the O₂ that reacts' in the key (own-lesson material word); all numbers pass · mcq4
127.0 g of copper is heated in a stream of O₂ and reacts completely to form copper(II) oxide, CuO: 2Cu + O₂ → 2CuO. Predict how the product's mass will compare with the copper's 127.0 g, and why.
Greater — the O₂ that reacts adds its own mass to the product. KEY
The same — mass is conserved, so the product must weigh exactly what the copper weighed.
Less — heating always drives some mass away as smoke.
Greater — hot substances weigh more than cold ones.
STO-005 · medium · application · seq 11 · authored (fresh reactant-needed-for-target instance; unwritten-1 given coefficient, subscripts as traps) · numeric
In 3Zn + N₂ → Zn₃N₂, how many moles of Zn are needed to make 5.0 mol of Zn₃N₂? Give your answer in moles to one decimal place.
Answer: 15.0 mol KEY
also accepted: 15.0 mol
wrongValueDiagnostics
[
{
"value": 1.7,
"misconception": "you turned the fraction upside down — the wanted substance Zn's coefficient 3 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Zn's coefficient 3 sits on top."
},
{
"value": 5.0,
"misconception": "you ignored the coefficients and matched the moles one-for-one — the ratio of Zn to Zn₃N₂ is 3:1",
"feedback": "This answer is wrong — you ignored the coefficients and matched the moles one-for-one — the ratio of Zn to Zn₃N₂ is 3:1."
},
{
"value": 8.0,
"misconception": "you added the coefficient instead of multiplying by the ratio",
"feedback": "This answer is wrong — you added the coefficient instead of multiplying by the ratio."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of Zn₃N₂ = 5.0 mol\nWrite down the equation:\nmoles of Zn = moles of Zn₃N₂ × (coefficient of Zn ÷ coefficient of Zn₃N₂)\nSubstitute in the values, and calculate:\nmoles of Zn = 5.0 × (3 ÷ 1)\nmoles of Zn = 15.0 mol"
}
STO-009 · medium · application · seq 12 · seed:U8-G1.demo_curated[1]/pool[25] (H2O2 chain) — structure kept, instance freshened to a corpus-free metal-halide chain (H2O2 6.8 g run is STO-013 WE0 corpus; 3.2/32.0/34.0 banned); re-instanced from 2K+I2->2KI after the parallel agent used KI (Pb(NO3)2+2KI) and 166.0 at row 19 · numeric
In 2Li + I₂ → 2LiI, 507.6 g of I₂ (M = 253.8 g/mol) reacts completely with Li. What mass of LiI forms? (M of LiI = 133.8 g/mol) Give your answer in grams to one decimal place.
Answer: 535.2 g KEY
also accepted: 535.2 g
wrongValueDiagnostics
[
{
"value": 267.6,
"misconception": "you skipped the mole-ratio step — 2.0 mol is the I₂'s moles, and twice as many moles of LiI form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 2.0 mol is the I₂'s moles, and twice as many moles of LiI form."
},
{
"value": 133.8,
"misconception": "you turned the fraction upside down — the wanted substance LiI's coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance LiI's coefficient 2 sits on top."
},
{
"value": 1015.2,
"misconception": "you used I₂'s molar mass in step 3 — the final step needs the WANTED substance LiI's molar mass",
"feedback": "This answer is wrong — you used I₂'s molar mass in step 3 — the final step needs the WANTED substance LiI's molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of I₂ = 507.6 g, M of I₂ = 253.8 g/mol, M of LiI = 133.8 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of I₂ = 507.6 / 253.8\nn of I₂ = 2.0 mol\nStep 2 — convert to moles of LiI with the mole ratio:\nmoles of LiI = moles of I₂ × (coefficient of LiI ÷ coefficient of I₂)\nmoles of LiI = 2.0 × (2 ÷ 1)\nmoles of LiI = 4.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of LiI = 4.0 × 133.8\nm of LiI = 535.2 g"
}
STO-011 · medium · application · seq 13 · authored (parallel numeric slot; fresh instance Sn + 2F2 -> SnF4, reactant-side total with an unwritten coefficient 1) · numeric
Sn reacts completely with F₂: Sn + 2F₂ → SnF₄. (M of Sn = 118.7 g/mol, M of F₂ = 38.0 g/mol.) What total mass of reactants does the balanced equation represent? Give your answer in grams to one decimal place.
Answer: 194.7 g KEY
also accepted: 194.7 g
wrongValueDiagnostics
[
{
"value": 156.7,
"misconception": "you ignored the coefficient 2 on F₂ — its mass is 2 × 38.0 g",
"feedback": "This answer is wrong — you ignored the coefficient 2 on F₂ — its mass is 2 × 38.0 g."
},
{
"value": 275.4,
"misconception": "you put the coefficient 2 on Sn instead of on F₂ — Sn has no written coefficient, so its coefficient is 1",
"feedback": "This answer is wrong — you put the coefficient 2 on Sn instead of on F₂ — Sn has no written coefficient, so its coefficient is 1."
},
{
"value": 158.7,
"misconception": "you added the coefficient instead of multiplying by it",
"feedback": "This answer is wrong — you added the coefficient instead of multiplying by it."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM of Sn = 118.7 g/mol, M of F₂ = 38.0 g/mol\nWrite down the equation:\nmass of each substance = coefficient × molar mass\nSubstitute in the values, and calculate:\ntotal mass of reactants = 1 × 118.7 + 2 × 38.0\ntotal mass of reactants = 194.7 g"
}
In 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O, 81.7 g of hexane, C₆H₁₄ (M = 86.0 g/mol), burns completely. How many moles of CO₂ form? Give your answer in moles to one decimal place.
Answer: 5.7 mol KEY
also accepted: 5.7 mol
wrongValueDiagnostics
[
{
"value": 0.158,
"misconception": "you turned the fraction upside down — the wanted substance CO₂'s coefficient 12 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance CO₂'s coefficient 12 sits on top."
},
{
"value": 0.95,
"misconception": "you stopped at step 1 — 0.95 mol is the hexane's moles, not the CO₂'s",
"feedback": "This answer is wrong — you stopped at step 1 — 0.95 mol is the hexane's moles, not the CO₂'s."
},
{
"value": 11.4,
"misconception": "you multiplied by CO₂'s coefficient alone instead of the coefficient ratio",
"feedback": "This answer is wrong — you multiplied by CO₂'s coefficient alone instead of the coefficient ratio."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of C₆H₁₄ = 81.7 g, M of C₆H₁₄ = 86.0 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of C₆H₁₄ = 81.7 / 86.0\nn of C₆H₁₄ = 0.95 mol\nStep 2 — convert to moles of CO₂ with the mole ratio:\nmoles of CO₂ = moles of C₆H₁₄ × (coefficient of CO₂ ÷ coefficient of C₆H₁₄)\nmoles of CO₂ = 0.95 × (12 ÷ 2)\nmoles of CO₂ = 5.7 mol"
}
STO-010 · medium · application · seq 15 · seed:U8-G1 STO-010 gateItems[0] (ethanol reverse chain) — instance freshened (46.0/44.0 corpus-banned; every fuel chain supplies a banned product M, so the row runs on a fresh metal-nitride surface 3Ba + N2 -> Ba3N2 with a 3:1 ratio) · numeric
In 3Ba + N₂ → Ba₃N₂, what mass of Ba must react completely with N₂ to make 439.9 g of Ba₃N₂? (M of Ba₃N₂ = 439.9 g/mol; M of Ba = 137.3 g/mol) Give your answer in grams to one decimal place.
Answer: 411.9 g KEY
also accepted: 411.9 g
wrongValueDiagnostics
[
{
"value": 137.3,
"misconception": "you skipped the mole-ratio step — 1.0 mol is the Ba₃N₂'s moles, and 3 mol of Ba are needed for each",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 1.0 mol is the Ba₃N₂'s moles, and 3 mol of Ba are needed for each."
},
{
"value": 45.8,
"misconception": "you turned the fraction upside down — the wanted substance Ba's coefficient 3 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Ba's coefficient 3 sits on top."
},
{
"value": 1319.7,
"misconception": "you used Ba₃N₂'s molar mass in step 3 — the final step needs the WANTED substance Ba's molar mass",
"feedback": "This answer is wrong — you used Ba₃N₂'s molar mass in step 3 — the final step needs the WANTED substance Ba's molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Ba₃N₂ = 439.9 g, M of Ba₃N₂ = 439.9 g/mol, M of Ba = 137.3 g/mol\nStep 1 — convert the target mass to moles:\nn = m / M\nn of Ba₃N₂ = 439.9 / 439.9\nn of Ba₃N₂ = 1.0 mol\nStep 2 — convert to moles of Ba with the mole ratio:\nmoles of Ba = moles of Ba₃N₂ × (coefficient of Ba ÷ coefficient of Ba₃N₂)\nmoles of Ba = 1.0 × (3 ÷ 1)\nmoles of Ba = 3.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of Ba = 3.0 × 137.3\nm of Ba = 411.9 g"
}
STO-008 · medium · application · seq 16 · fully authored (F3 reverse pairing; fresh instance 3Ca + N2 -> Ca3N2, unwritten coefficient 1 on the given substance; withheld M(Ca3N2)=148.3 named in the wrong-M misconception) · numeric
In 3Ca + N₂ → Ca₃N₂, what mass of Ca is needed to make 3.0 mol of Ca₃N₂? (M of Ca = 40.1 g/mol) Give your answer in grams to one decimal place.
Answer: 360.9 g KEY
also accepted: 360.9 g
wrongValueDiagnostics
[
{
"value": 120.3,
"misconception": "you skipped the mole-ratio step — 3.0 mol is the Ca₃N₂'s moles, and 3 mol of Ca are needed for each",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 3.0 mol is the Ca₃N₂'s moles, and 3 mol of Ca are needed for each."
},
{
"value": 40.1,
"misconception": "you turned the fraction upside down — the wanted substance Ca's coefficient 3 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Ca's coefficient 3 sits on top."
},
{
"value": 1334.7,
"misconception": "you used Ca₃N₂'s molar mass (148.3 g/mol) — step 2 needs the WANTED substance Ca's molar mass",
"feedback": "This answer is wrong — you used Ca₃N₂'s molar mass (148.3 g/mol) — step 2 needs the WANTED substance Ca's molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of Ca₃N₂ = 3.0 mol, M of Ca = 40.1 g/mol\nStep 1 — convert to moles of Ca with the mole ratio:\nmoles of Ca = moles of Ca₃N₂ × (coefficient of Ca ÷ coefficient of Ca₃N₂)\nmoles of Ca = 3.0 × (3 ÷ 1)\nmoles of Ca = 9.0 mol\nStep 2 — convert the moles to mass:\nm = n × M\nm of Ca = 9.0 × 40.1\nm of Ca = 360.9 g"
}
STO-011 · hard · transfer · seq 17 · authored (fresh audit per blueprint case rotation; error class: reactant coefficient ignored; wave-5 fresh instance 2Sb + 3I₂ → 2SbI₃) · mcq4
For 2Sb + 3I₂ → 2SbI₃, a student totals the two sides and gets 883.2 g of reactants but 1005.0 g of products. (M of Sb = 121.8 g/mol, M of I₂ = 253.8 g/mol, M of SbI₃ = 502.5 g/mol.) What went wrong?
The coefficient 2 on Sb was ignored — the reactant side is 2 × 121.8 + 3 × 253.8 = 1005.0 g. KEY
The product total is wrong — the SbI₃ should carry a coefficient of 1, giving 502.5 g.
The equation itself is unbalanced, so the totals cannot match.
Nothing went wrong — a synthesis reaction always gains mass as the product forms.
STO-013 · hard · transfer · seq 18 · authored to seed:U8-G1.pool[37] class (STO-013 gateItems[1], step-1 self-consistent wrong-M) — instance freshened (2CO + O2 is STO-011 fluency corpus; 28.0/44.0 banned) · mcq4
For 4Cr + 3O₂ → 2Cr₂O₃, a student calculates the mass of Cr₂O₃ formed from 197.6 g of Cr. (M of Cr = 52.0 g/mol, M of Cr₂O₃ = 152.0 g/mol.)
Step 1: n of Cr = 197.6 / 152.0 = 1.3 mol
Step 2: moles of Cr₂O₃ = 1.3 × 2/4 = 0.65 mol
Step 3: m of Cr₂O₃ = 0.65 × 152.0 = 98.8 g
Which step contains the error?
Step 1 KEY
Step 2
Step 3
No step contains an error
STO-010 · hard · transfer · seq 19 · fully authored per blueprint (production-planning framing, awkward 3:1 pair, both Ms supplied; wrongs: skipped ratio / wrong-partner coefficient / wrong-M — inverted dropped for this instance because 2.0 × (1/3) does not divide exactly; wave-5 fresh instance 3Co + 2O₂ → Co₃O₄) · numeric
In 3Co + 2O₂ → Co₃O₄, a battery-materials plant must produce 481.4 g of Co₃O₄ for one order. What mass of Co must burn completely to make it? (M of Co₃O₄ = 240.7 g/mol; M of Co = 58.9 g/mol) Give your answer in grams to one decimal place.
Answer: 353.4 g KEY
also accepted: 353.4 g
wrongValueDiagnostics
[
{
"value": 117.8,
"misconception": "you skipped the mole-ratio step — 2.0 mol is the Co₃O₄'s moles, and three times as many moles of Co are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 2.0 mol is the Co₃O₄'s moles, and three times as many moles of Co are needed."
},
{
"value": 235.6,
"misconception": "you used O₂'s coefficient 2 in the ratio — the wanted substance is Co, whose coefficient is 3",
"feedback": "This answer is wrong — you used O₂'s coefficient 2 in the ratio — the wanted substance is Co, whose coefficient is 3."
},
{
"value": 1444.2,
"misconception": "you used Co₃O₄'s molar mass in step 3 — the final step needs the WANTED substance Co's molar mass",
"feedback": "This answer is wrong — you used Co₃O₄'s molar mass in step 3 — the final step needs the WANTED substance Co's molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Co₃O₄ = 481.4 g, M of Co₃O₄ = 240.7 g/mol, M of Co = 58.9 g/mol\nStep 1 — convert the target mass to moles:\nn = m / M\nn of Co₃O₄ = 481.4 / 240.7\nn of Co₃O₄ = 2.0 mol\nStep 2 — convert to moles of Co with the mole ratio:\nmoles of Co = moles of Co₃O₄ × (coefficient of Co ÷ coefficient of Co₃O₄)\nmoles of Co = 2.0 × (3 ÷ 1)\nmoles of Co = 6.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of Co = 6.0 × 58.9\nm of Co = 353.4 g"
}
STO-009 · hard · transfer · seq 20 · fully authored per blueprint (capstone terminal slot: full forward chain, awkward 5:2 pair, industrial transfer surface — calciothermic V production; both Ms enrolled a8 atomic constants; wave-10 sol realism fix: 4K+O2->2K2O dropped (K combustion favors the superoxide); Al(OH)3-calcination and FeS2-roasting repicks census-killed) · numeric
In a metal refinery, V₂O₅ + 5Ca → 2V + 5CaO, 401.0 g of Ca (M = 40.1 g/mol) reacts completely with V₂O₅. (M of V = 50.9 g/mol) What mass of V forms? Give your answer in grams to one decimal place.
Answer: 203.6 g KEY
also accepted: 203.6 g
wrongValueDiagnostics
[
{
"value": 509.0,
"misconception": "you skipped the mole-ratio step — 10.0 mol is the Ca's moles, and fewer moles of V form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 10.0 mol is the Ca's moles, and fewer moles of V form."
},
{
"value": 1272.5,
"misconception": "you turned the fraction upside down — the wanted substance V's coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance V's coefficient 2 sits on top."
},
{
"value": 160.4,
"misconception": "you used Ca's molar mass in step 3 — the final step needs the WANTED substance V's molar mass",
"feedback": "This answer is wrong — you used Ca's molar mass in step 3 — the final step needs the WANTED substance V's molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Ca = 401.0 g, M of Ca = 40.1 g/mol, M of V = 50.9 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Ca = 401.0 / 40.1\nn of Ca = 10.0 mol\nStep 2 — convert to moles of V with the mole ratio:\nmoles of V = moles of Ca × (coefficient of V ÷ coefficient of Ca)\nmoles of V = 10.0 × (2 ÷ 5)\nmoles of V = 4.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of V = 4.0 × 50.9\nm of V = 203.6 g"
}
STO-005 · medium · application · seq 21 · authored (FRQ1 per composition.frq_split_rationale: STO-005 mole-to-mole in BOTH roles + taught sanity check; mole-only, no molar masses per the STO-007+ boundary; fresh reaction + numbers per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
Mg reacts with HCl according to the balanced equation Mg + 2HCl → MgCl₂ + H₂. Work in moles throughout — no molar masses are needed.
Mg reacts with HCl according to the balanced equation Mg + 2HCl → MgCl₂ + H₂. Work in moles throughout — no molar masses are needed.
(a) 8.0 mol of HCl reacts completely with Mg. Calculate the moles of MgCl₂ that form. Show your working: state the given moles, write the mole-ratio equation with both substances named, substitute, and calculate.
(b) Calculate the moles of HCl needed to produce 3.0 mol of H₂. Show the same working.
(c) For part (a), apply the lesson's sanity check without recalculating: compare the coefficient of the wanted substance with the coefficient of the given substance, state whether the answer should come out smaller or larger than the given moles, and confirm that your part (a) answer passes the check.
Part a. 8.0 mol of HCl reacts completely with Mg. Calculate the moles of MgCl₂ that form. Show your working: state the given moles, write the mole-ratio equation with both substances named, substitute, and calculate.
Model: moles of HCl = 8.0 mol
moles of MgCl2 = moles of HCl × (coefficient of MgCl2 ÷ coefficient of HCl)
moles of MgCl2 = 8.0 × (1 ÷ 2)
moles of MgCl2 = 4.0 mol
Part b. Calculate the moles of HCl needed to produce 3.0 mol of H₂. Show the same working.
Model: moles of hydrogen = 3.0 mol (the target)
moles of HCl = moles of hydrogen × (coefficient of HCl ÷ coefficient of hydrogen)
moles of HCl = 3.0 × (2 ÷ 1)
moles of HCl = 6.0 mol
Part c. For part (a), apply the lesson's sanity check without recalculating: compare the coefficient of the wanted substance with the coefficient of the given substance, state whether the answer should come out smaller or larger than the given moles, and confirm that your part (a) answer passes the check.
Model: The taught check says: when the wanted coefficient is smaller than the given one, the answer comes out smaller than the given moles. In part (a) the wanted substance MgCl2 has coefficient 1 and the given substance HCl has coefficient 2 — the wanted coefficient is smaller, so the answer should come out smaller than 8.0 mol. The answer 4.0 mol is smaller than 8.0 mol, so it passes the check.
Full model answer. (a)
moles of HCl = 8.0 mol
moles of MgCl2 = moles of HCl × (coefficient of MgCl2 ÷ coefficient of HCl)
moles of MgCl2 = 8.0 × (1 ÷ 2)
moles of MgCl2 = 4.0 mol
(b)
moles of hydrogen = 3.0 mol (the target)
moles of HCl = moles of hydrogen × (coefficient of HCl ÷ coefficient of hydrogen)
moles of HCl = 3.0 × (2 ÷ 1)
moles of HCl = 6.0 mol
(c)
The taught check says: when the wanted coefficient is smaller than the given one, the answer comes out smaller than the given moles. In part (a) the wanted substance MgCl2 has coefficient 1 and the given substance HCl has coefficient 2 — the wanted coefficient is smaller, so the answer should come out smaller than 8.0 mol. The answer 4.0 mol is smaller than 8.0 mol, so it passes the check.
STO-009 · hard · application · seq 22 · authored (FRQ2 per composition.frq_split_rationale: STO-009 capstone chain end-to-end — plan + full mass-to-mass run + step-3 molar-mass rule; forward reactant→product only per the STO-010 boundary; both Ms supplied, corpus-free; fresh reaction + masses per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
Cu is heated with S to form Cu₂S as the balanced equation shows 2Cu + S → Cu₂S. 254.0 g of Cu reacts completely. (M of Cu = 63.5 g/mol; M of Cu₂S = 159.1 g/mol.)
Cu is heated with S to form Cu₂S as the balanced equation shows 2Cu + S → Cu₂S. 254.0 g of Cu reacts completely. (M of Cu = 63.5 g/mol; M of Cu₂S = 159.1 g/mol.)
(a) Before calculating, state the three-step plan for working out the mass of Cu₂S that forms, naming the tool each step uses.
(b) Carry out the calculation, showing the working for every step, and give the mass of Cu₂S that forms.
(c) State which substance's molar mass belongs in step 3, and why.
Part a. Before calculating, state the three-step plan for working out the mass of Cu₂S that forms, naming the tool each step uses.
Model: grams of Cu → moles of Cu (n = m/M) → moles of the copper sulfide (mole-ratio equation) → grams of the copper sulfide (m = n × M).
Part b. Carry out the calculation, showing the working for every step, and give the mass of Cu₂S that forms.
Model: m of Cu = 254.0 g; M of Cu = 63.5 g/mol; M of the copper sulfide = 159.1 g/mol
Step 1 — convert the mass to moles:
n = m / M
n of Cu = 254.0 / 63.5
n of Cu = 4.0 mol
Step 2 — convert to moles of the copper sulfide with the mole ratio:
moles of the copper sulfide = moles of Cu × (coefficient of the copper sulfide ÷ coefficient of Cu)
moles of the copper sulfide = 4.0 × (1 ÷ 2)
moles of the copper sulfide = 2.0 mol
Step 3 — convert the moles to mass:
m = n × M
m of the copper sulfide = 2.0 × 159.1
m of the copper sulfide = 318.2 g
Part c. State which substance's molar mass belongs in step 3, and why.
Model: Step 3 must use the WANTED substance's molar mass — the copper sulfide's, 159.1 g/mol — because step 3 converts the wanted substance's moles into its mass with m = n × M. The given substance's molar mass (Cu's 63.5 g/mol) is used only in step 1 and never appears in step 3.
Full model answer. (a)
grams of Cu → moles of Cu (n = m/M) → moles of the copper sulfide (mole-ratio equation) → grams of the copper sulfide (m = n × M).
(b)
m of Cu = 254.0 g; M of Cu = 63.5 g/mol; M of the copper sulfide = 159.1 g/mol
Step 1 — convert the mass to moles:
n = m / M
n of Cu = 254.0 / 63.5
n of Cu = 4.0 mol
Step 2 — convert to moles of the copper sulfide with the mole ratio:
moles of the copper sulfide = moles of Cu × (coefficient of the copper sulfide ÷ coefficient of Cu)
moles of the copper sulfide = 4.0 × (1 ÷ 2)
moles of the copper sulfide = 2.0 mol
Step 3 — convert the moles to mass:
m = n × M
m of the copper sulfide = 2.0 × 159.1
m of the copper sulfide = 318.2 g
(c)
Step 3 must use the WANTED substance's molar mass — the copper sulfide's, 159.1 g/mol — because step 3 converts the wanted substance's moles into its mass with m = n × M. The given substance's molar mass (Cu's 63.5 g/mol) is used only in step 1 and never appears in step 3.
STO-013 · hard · transfer · seq 23 · authored (FRQ3 per composition.frq_split_rationale: STO-013 inspection PRODUCTION — one planted error; B-form class = step-3 wrong-M (third substance's M, MgCl2's, supplied in-stem so the slip is checkable); fresh reaction + numbers per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
For Mg + CuCl₂ → MgCl₂ + Cu, a student calculates the mass of Cu formed when 403.5 g of CuCl₂ reacts completely with Mg in solution. (M of CuCl₂ = 134.5 g/mol; M of Cu = 63.5 g/mol; M of MgCl₂ = 95.3 g/mol.)
Step 1: n of CuCl₂ = 403.5 / 134.5 = 3.0 mol
Step 2: moles of Cu = 3.0 × (1 ÷ 1) = 3.0 mol
Step 3: m of Cu = 3.0 × 95.3 = 285.9 g
Exactly one step contains an error.
For Mg + CuCl₂ → MgCl₂ + Cu, a student calculates the mass of Cu formed when 403.5 g of CuCl₂ reacts completely with Mg in solution. (M of CuCl₂ = 134.5 g/mol; M of Cu = 63.5 g/mol; M of MgCl₂ = 95.3 g/mol.)
Step 1: n of CuCl₂ = 403.5 / 134.5 = 3.0 mol
Step 2: moles of Cu = 3.0 × (1 ÷ 1) = 3.0 mol
Step 3: m of Cu = 3.0 × 95.3 = 285.9 g
Exactly one step contains an error.
(a) Name the step that contains the error, and state what is wrong with it.
(b) Correct the erroneous step and give the corrected mass of Cu, showing the working.
(c) A classmate claims step 2 must also be wrong because the final mass came out wrong. Using the lesson's rule that an error lives where the wrong number was made, state which step made the wrong number and whether any step in this calculation merely inherited one.
Part a. Name the step that contains the error, and state what is wrong with it.
Model: The error is in step 3. Checking each step: step 1 divides 403.5 g by CuCl2's own molar mass, 134.5 g/mol — correct. Step 2 reads the ratio coefficient of Cu ÷ coefficient of CuCl2 = 1 ÷ 1 from the balanced equation — correct. Step 3 must use the WANTED substance's molar mass, Cu's 63.5 g/mol, but the student used MgCl2's 95.3 g/mol.
Part b. Correct the erroneous step and give the corrected mass of Cu, showing the working.
Model: Corrected step 3:
moles of Cu = 3.0 mol
m = n × M
m of Cu = 3.0 × 63.5
m of Cu = 190.5 g
Part c. A classmate claims step 2 must also be wrong because the final mass came out wrong. Using the lesson's rule that an error lives where the wrong number was made, state which step made the wrong number and whether any step in this calculation merely inherited one.
Model: The wrong number was made in step 3 itself: steps 1 and 2 both made correct numbers, so the classmate is wrong about step 2. No step in this calculation merely inherited a wrong number — nothing comes after step 3, so the rule that a step inheriting a wrong number is not itself the error has nothing to protect here. The error lives in step 3, where the wrong number was made.
Full model answer. (a)
The error is in step 3. Checking each step: step 1 divides 403.5 g by CuCl2's own molar mass, 134.5 g/mol — correct. Step 2 reads the ratio coefficient of Cu ÷ coefficient of CuCl2 = 1 ÷ 1 from the balanced equation — correct. Step 3 must use the WANTED substance's molar mass, Cu's 63.5 g/mol, but the student used MgCl2's 95.3 g/mol.
(b)
Corrected step 3:
moles of Cu = 3.0 mol
m = n × M
m of Cu = 3.0 × 63.5
m of Cu = 190.5 g
(c)
The wrong number was made in step 3 itself: steps 1 and 2 both made correct numbers, so the classmate is wrong about step 2. No step in this calculation merely inherited a wrong number — nothing comes after step 3, so the rule that a step inheriting a wrong number is not itself the error has nothing to protect here. The error lives in step 3, where the wrong number was made.
In Sn + 2Cl₂ → SnCl₄, 2 molecules of Cl₂ react for every 1 atom of Sn. Why does the same two-to-one ratio hold for moles of Cl₂ and moles of Sn?
Because a mole is the same fixed count for every substance, so scaling every substance up by that same count leaves the ratio unchanged. KEY
Because 2 g of Cl₂ react for every 1 g of Sn, and whatever holds for the gram ratio carries straight over to the mole ratio.
Because Cl₂ and Sn have the same molar mass, so equal counts of them always weigh the same.
Because the mole is chosen fresh for each reaction so that the coefficients always describe it correctly.
STO-001 · easy · recall · seq 2 · authored (fresh burn with both coefficients written, coprime as asked; wave-5 fresh instance 4Al + 3O₂ → 2Al₂O₃) · text
In the balanced equation 4Al + 3O₂ → 2Al₂O₃, what is the mole ratio of Al to O₂? Write the ratio with a colon, such as 4:1.
Answer: 4:3 KEY
also accepted: 4 : 3, 4 to 3
wrongValueDiagnostics
[
{
"value": "3:4",
"misconception": "you reversed the order — the question asks Al to O₂, so Al's coefficient comes first",
"feedback": "This answer is wrong — you reversed the order. The question asks Al to O₂, so Al's coefficient 4 comes first: 4:3."
},
{
"value": "4:2",
"misconception": "you took Al₂O₃'s coefficient for O₂ — the second number must belong to the substance named second",
"feedback": "This answer is wrong — you took Al₂O₃'s coefficient 2 for O₂. The second number must belong to the substance named second, O₂, whose coefficient is 3: 4:3."
},
{
"value": "1:1",
"misconception": "you ignored the coefficients — the mole ratio is read from the written coefficients, 4 for Al and 3 for O₂",
"feedback": "This answer is wrong — you ignored the coefficients. The mole ratio comes from the written coefficients: 4 for Al and 3 for O₂, so the ratio is 4:3."
}
]
feedback
{
"keyExplanation": "The coefficient of Al is 4, and the coefficient of O₂ is 3. Take the coefficients in the order asked: Al first, O₂ second. The mole ratio of Al to O₂ is 4:3.",
"anticipatedWrong": [
{
"value": "3:4",
"misconception": "you reversed the order — the question asks Al to O₂, so Al's coefficient comes first",
"feedback": "This answer is wrong — you reversed the order. The question asks Al to O₂, so Al's coefficient 4 comes first: 4:3."
},
{
"value": "4:2",
"misconception": "you took Al₂O₃'s coefficient for O₂ — the second number must belong to the substance named second",
"feedback": "This answer is wrong — you took Al₂O₃'s coefficient 2 for O₂. The second number must belong to the substance named second, O₂, whose coefficient is 3: 4:3."
},
{
"value": "1:1",
"misconception": "you ignored the coefficients — the mole ratio is read from the written coefficients, 4 for Al and 3 for O₂",
"feedback": "This answer is wrong — you ignored the coefficients. The mole ratio comes from the written coefficients: 4 for Al and 3 for O₂, so the ratio is 4:3."
}
]
}
STO-003 · easy · recall · seq 3 · authored (same miscounted-atoms reason read against a fresh unbalanced-vs-balanced pair, kept qualitative; wave-6 fresh instance S + F₂ → SF₆ vs S + 3F₂ → SF₆; foils = small-amounts-ok / direction / fixed-factor) · mcq4
A worksheet shows the unbalanced equation S + F₂ → SF₆ next to the balanced version S + 3F₂ → SF₆. Why is the unbalanced version unusable for mole ratios?
Its coefficients miscount atoms — F atoms would have to appear from nowhere, and a reaction only rearranges atoms. KEY
Its coefficients are fine for small amounts — an unbalanced equation only fails when large amounts react.
Its coefficients put the reactants and the product on the wrong sides of the arrow, reversing the reaction.
Its coefficients come out exactly half of the true values, so the true ratio is found by doubling them.
STO-006 · easy · recall · seq 4 · authored (step-order rotation per blueprint — which-step-remains read, instance-free, zero arithmetic per JM-P1a) · mcq4
In a mass-to-mass calculation, a student has just used the mole ratio and now holds the moles of the wanted substance. Which step remains?
Convert those moles to the wanted substance's mass with m = n × M. KEY
Convert those moles back to moles of the given substance to confirm the ratio.
Convert the given substance's mass to its moles with n = m/M.
Apply the mole ratio a second time to turn the moles into grams.
In 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O, 3.0 mol of C₃H₆ burns completely. How many moles of CO₂ form? Give your answer in moles to one decimal place.
Answer: 9.0 mol KEY
also accepted: 9.0 mol
wrongValueDiagnostics
[
{
"value": 1.0,
"misconception": "you turned the fraction upside down — the wanted substance CO₂'s coefficient 6 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance CO₂'s coefficient 6 sits on top."
},
{
"value": 3.0,
"misconception": "you ignored the coefficients and matched the moles one-for-one — the ratio of CO₂ to C₃H₆ is 6:2",
"feedback": "This answer is wrong — you ignored the coefficients and matched the moles one-for-one — the ratio of CO₂ to C₃H₆ is 6:2."
},
{
"value": 18.0,
"misconception": "you multiplied by CO₂'s coefficient alone instead of the coefficient ratio",
"feedback": "This answer is wrong — you multiplied by CO₂'s coefficient alone instead of the coefficient ratio."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of C₃H₆ = 3.0 mol\nWrite down the equation:\nmoles of CO₂ = moles of C₃H₆ × (coefficient of CO₂ ÷ coefficient of C₃H₆)\nSubstitute in the values, and calculate:\nmoles of CO₂ = 3.0 × (6 ÷ 2)\nmoles of CO₂ = 9.0 mol"
}
STO-001 · medium · application · seq 7 · authored (parallel text slot; fresh unwritten-1 pair with the inside-subscript trap on the asked substance N₂) · text
In the balanced equation 6Li + N₂ → 2Li₃N, what is the mole ratio of N₂ to Li? Write the ratio with a colon, such as 4:1.
Answer: 1:6 KEY
also accepted: 1 : 6, 1 to 6
wrongValueDiagnostics
[
{
"value": "2:6",
"misconception": "you used the subscript 2 inside N₂ as its coefficient — N₂ has no written coefficient, so its coefficient is 1",
"feedback": "This answer is wrong — you used the subscript 2 inside N₂ as its coefficient — N₂ has no written coefficient, so its coefficient is 1."
},
{
"value": "6:1",
"misconception": "you reversed the order — the question asks N₂ to Li, so N₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order — the question asks N₂ to Li, so N₂'s coefficient comes first."
},
{
"value": "1:2",
"misconception": "you took Li₃N's coefficient for Li — the second number must belong to Li, whose coefficient is 6",
"feedback": "This answer is wrong — you took Li₃N's coefficient for Li — the second number must belong to Li, whose coefficient is 6."
}
]
feedback
{
"keyExplanation": "N₂ has no written coefficient, so its coefficient is 1; the coefficient of Li is 6. The subscript 2 inside N₂ counts atoms in one molecule and plays no part in the mole ratio. The mole ratio of N₂ to Li is 1:6.",
"anticipatedWrong": [
{
"value": "2:6",
"misconception": "you used the subscript 2 inside N₂ as its coefficient — N₂ has no written coefficient, so its coefficient is 1",
"feedback": "This answer is wrong — you used the subscript 2 inside N₂ as its coefficient — N₂ has no written coefficient, so its coefficient is 1."
},
{
"value": "6:1",
"misconception": "you reversed the order — the question asks N₂ to Li, so N₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order — the question asks N₂ to Li, so N₂'s coefficient comes first."
},
{
"value": "1:2",
"misconception": "you took Li₃N's coefficient for Li — the second number must belong to Li, whose coefficient is 6",
"feedback": "This answer is wrong — you took Li₃N's coefficient for Li — the second number must belong to Li, whose coefficient is 6."
}
]
}
STO-007 · medium · application · seq 8 · authored (parallel numeric slot; fresh metal + acid surface per the case rotation, spoken in formulas; only the given substance's M supplied) · numeric
In Zn + 2HCl → ZnCl₂ + H₂, 163.5 g of Zn (M = 65.4 g/mol) reacts completely with HCl. How many moles of HCl are used up? Give your answer in moles to 2 significant figures.
Answer: 5.0 mol KEY
also accepted: 5.0 mol
wrongValueDiagnostics
[
{
"value": 1.25,
"misconception": "you turned the fraction upside down — the wanted substance HCl's coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance HCl's coefficient 2 sits on top."
},
{
"value": 2.5,
"misconception": "you skipped the mole-ratio step — 2.5 mol is the Zn's moles, not the HCl's",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 2.5 mol is the Zn's moles, not the HCl's."
},
{
"value": 327.0,
"misconception": "you applied the mole ratio to the mass instead of converting to moles first",
"feedback": "This answer is wrong — you applied the mole ratio to the mass instead of converting to moles first."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Zn = 163.5 g, M of Zn = 65.4 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Zn = 163.5 / 65.4\nn of Zn = 2.5 mol\nStep 2 — convert to moles of HCl with the mole ratio:\nmoles of HCl = moles of Zn × (coefficient of HCl ÷ coefficient of Zn)\nmoles of HCl = 2.5 × (2 ÷ 1)\nmoles of HCl = 5.0 mol"
}
STO-008 · medium · application · seq 9 · seed:U8-G1.pool[23] (STO-008 gateItems[2], CH4 2:1 read) — 2:1-class shape kept, instance freshened to a metal-sulfide surface (CH4 equation + M(H2O)=18.0 corpus-dead); re-instanced from Ag2S after the parallel agent homed 2Ag+S->Ag2S at row 18 · numeric
In Ti + 2S → TiS₂, 6.0 mol of S reacts completely with Ti. What mass of TiS₂ forms? (M of TiS₂ = 112.1 g/mol) Give your answer in grams to one decimal place.
Answer: 336.3 g KEY
also accepted: 336.3 g
wrongValueDiagnostics
[
{
"value": 672.6,
"misconception": "you skipped the mole-ratio step — 6.0 mol is the S's moles, and only half as many moles of TiS₂ form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 6.0 mol is the S's moles, and only half as many moles of TiS₂ form."
},
{
"value": 1345.2,
"misconception": "you turned the fraction upside down — the wanted substance TiS₂'s coefficient 1 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance TiS₂'s coefficient 1 sits on top."
},
{
"value": 96.3,
"misconception": "you used S's molar mass (32.1 g/mol) — step 2 needs the WANTED substance TiS₂'s molar mass",
"feedback": "This answer is wrong — you used S's molar mass (32.1 g/mol) — step 2 needs the WANTED substance TiS₂'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of S = 6.0 mol, M of TiS₂ = 112.1 g/mol\nStep 1 — convert to moles of TiS₂ with the mole ratio:\nmoles of TiS₂ = moles of S × (coefficient of TiS₂ ÷ coefficient of S)\nmoles of TiS₂ = 6.0 × (1 ÷ 2)\nmoles of TiS₂ = 3.0 mol\nStep 2 — convert the moles to mass:\nm = n × M\nm of TiS₂ = 3.0 × 112.1\nm of TiS₂ = 336.3 g"
}
STO-012 · medium · application · seq 10 · seed:U8-G1.pool[35] (STO-012 gateItems[2], propane-stove collected-products) — structure and misconception space kept, instance freshened to an acetylene torch (propane is a corpus-wide material word; acetylene HOME seq 10-C per blueprint); seed key's 'Oxygen' -> 'O₂' (own-lesson material word) · mcq4
A welding torch burns acetylene, C₂H₂, from a pressurized tank. A chemist collects ALL of the CO₂ and H₂O the flame produces and finds their combined mass is greater than the mass of acetylene used. Why?
The O₂ that fed the flame reacted too, and its mass is inside the CO₂ and H₂O. KEY
The flame converted some of its heat and light energy into extra mass.
Acetylene weighs more once it leaves the pressurized tank.
The measurement of the acetylene used must be wrong.
STO-005 · medium · application · seq 11 · authored (fresh other-reactant-needed instance, 3:2 awkward ratio; Au = classroom-safe metal-halide family per the allocation's restricted-instance rule; wave-6 fresh instance 2Au + 3Cl₂ → 2AuCl₃) · numeric
In 2Au + 3Cl₂ → 2AuCl₃, how many moles of Cl₂ are needed to react completely with 4.0 mol of Au? Give your answer in moles to one decimal place.
Answer: 6.0 mol KEY
also accepted: 6.0 mol
wrongValueDiagnostics
[
{
"value": 2.7,
"misconception": "you turned the fraction upside down — the wanted substance Cl₂'s coefficient 3 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Cl₂'s coefficient 3 sits on top."
},
{
"value": 4.0,
"misconception": "you ignored the coefficients and matched the moles one-for-one — the ratio of Cl₂ to Au is 3:2",
"feedback": "This answer is wrong — you ignored the coefficients and matched the moles one-for-one — the ratio of Cl₂ to Au is 3:2."
},
{
"value": 12.0,
"misconception": "you forgot Au's coefficient 2 on the bottom of the fraction",
"feedback": "This answer is wrong — you forgot Au's coefficient 2 on the bottom of the fraction."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of Au = 4.0 mol\nWrite down the equation:\nmoles of Cl₂ = moles of Au × (coefficient of Cl₂ ÷ coefficient of Au)\nSubstitute in the values, and calculate:\nmoles of Cl₂ = 4.0 × (3 ÷ 2)\nmoles of Cl₂ = 6.0 mol"
}
STO-009 · medium · application · seq 12 · seed:U8-G1.demo_curated[2]/pool[26] (N2->NO chain) — structure kept, instance freshened to a corpus-free metal-nitride chain (N2+O2->2NO is STO-002 WE0 corpus; 14.0/28.0/30.0 banned); Sr3N2 equation is bank-fresh (existing rows use only 2Sr+O2) · numeric
In 3Sr + N₂ → Sr₃N₂, 131.4 g of Sr (M = 87.6 g/mol) reacts completely with N₂. What mass of Sr₃N₂ forms? (M of Sr₃N₂ = 290.8 g/mol) Give your answer in grams to one decimal place.
Answer: 145.4 g KEY
also accepted: 145.4 g
wrongValueDiagnostics
[
{
"value": 436.2,
"misconception": "you skipped the mole-ratio step — 1.5 mol is the Sr's moles, and only a third as many moles of Sr₃N₂ form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 1.5 mol is the Sr's moles, and only a third as many moles of Sr₃N₂ form."
},
{
"value": 1308.6,
"misconception": "you turned the fraction upside down — the wanted substance Sr₃N₂'s coefficient 1 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Sr₃N₂'s coefficient 1 sits on top."
},
{
"value": 43.8,
"misconception": "you used Sr's molar mass in step 3 — the final step needs the WANTED substance Sr₃N₂'s molar mass",
"feedback": "This answer is wrong — you used Sr's molar mass in step 3 — the final step needs the WANTED substance Sr₃N₂'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Sr = 131.4 g, M of Sr = 87.6 g/mol, M of Sr₃N₂ = 290.8 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Sr = 131.4 / 87.6\nn of Sr = 1.5 mol\nStep 2 — convert to moles of Sr₃N₂ with the mole ratio:\nmoles of Sr₃N₂ = moles of Sr × (coefficient of Sr₃N₂ ÷ coefficient of Sr)\nmoles of Sr₃N₂ = 1.5 × (1 ÷ 3)\nmoles of Sr₃N₂ = 0.50 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of Sr₃N₂ = 0.50 × 290.8\nm of Sr₃N₂ = 145.4 g"
}
STO-011 · medium · application · seq 13 · authored (parallel numeric slot; fresh instance 2Al + 3I2 -> 2AlI3 — Al spoken as SYMBOL, aluminum-oxygen corpus instance avoided) · numeric
Al reacts completely with I₂: 2Al + 3I₂ → 2AlI₃. (M of AlI₃ = 407.7 g/mol.) What total mass of products does the balanced equation represent? Give your answer in grams to one decimal place.
Answer: 815.4 g KEY
also accepted: 815.4 g
wrongValueDiagnostics
[
{
"value": 407.7,
"misconception": "you ignored the coefficient 2 on AlI₃ — its mass is 2 × 407.7 g",
"feedback": "This answer is wrong — you ignored the coefficient 2 on AlI₃ — its mass is 2 × 407.7 g."
},
{
"value": 1223.1,
"misconception": "you multiplied by 3, the I₂ coefficient, instead of by AlI₃'s own coefficient 2",
"feedback": "This answer is wrong — you multiplied by 3, the I₂ coefficient, instead of by AlI₃'s own coefficient 2."
},
{
"value": 409.7,
"misconception": "you added the coefficient instead of multiplying by it",
"feedback": "This answer is wrong — you added the coefficient instead of multiplying by it."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM of AlI₃ = 407.7 g/mol, coefficient of AlI₃ = 2\nWrite down the equation:\nmass of each substance = coefficient × molar mass\nSubstitute in the values, and calculate:\ntotal mass of products = 2 × 407.7\ntotal mass of products = 815.4 g"
}
In 2C₄H₉OH + 12O₂ → 8CO₂ + 10H₂O, 62.9 g of butanol, C₄H₉OH (M = 74.0 g/mol), burns completely. How many moles of CO₂ form? Give your answer in moles to one decimal place.
Answer: 3.4 mol KEY
also accepted: 3.4 mol
wrongValueDiagnostics
[
{
"value": 0.21,
"misconception": "you turned the fraction upside down — the wanted substance CO₂'s coefficient 8 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance CO₂'s coefficient 8 sits on top."
},
{
"value": 0.85,
"misconception": "you stopped at step 1 — 0.85 mol is the butanol's moles, not the CO₂'s",
"feedback": "This answer is wrong — you stopped at step 1 — 0.85 mol is the butanol's moles, not the CO₂'s."
},
{
"value": 251.6,
"misconception": "you applied the mole ratio to the mass instead of converting to moles first",
"feedback": "This answer is wrong — you applied the mole ratio to the mass instead of converting to moles first."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of C₄H₉OH = 62.9 g, M of C₄H₉OH = 74.0 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of C₄H₉OH = 62.9 / 74.0\nn of C₄H₉OH = 0.85 mol\nStep 2 — convert to moles of CO₂ with the mole ratio:\nmoles of CO₂ = moles of C₄H₉OH × (coefficient of CO₂ ÷ coefficient of C₄H₉OH)\nmoles of CO₂ = 0.85 × (8 ÷ 2)\nmoles of CO₂ = 3.4 mol"
}
In Mn + 2HBr → MnBr₂ + H₂, what mass of HBr must react completely with manganese to make 429.4 g of MnBr₂? (M of MnBr₂ = 214.7 g/mol; M of HBr = 80.9 g/mol) Give your answer in grams to one decimal place.
Answer: 323.6 g KEY
also accepted: 323.6 g
wrongValueDiagnostics
[
{
"value": 161.8,
"misconception": "you skipped the mole-ratio step — 2.0 mol is the MnBr₂'s moles, and twice as many moles of HBr are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 2.0 mol is the MnBr₂'s moles, and twice as many moles of HBr are needed."
},
{
"value": 80.9,
"misconception": "you turned the fraction upside down — the wanted substance HBr's coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance HBr's coefficient 2 sits on top."
},
{
"value": 858.8,
"misconception": "you used MnBr₂'s molar mass in step 3 — the final step needs the WANTED substance HBr's molar mass",
"feedback": "This answer is wrong — you used MnBr₂'s molar mass in step 3 — the final step needs the WANTED substance HBr's molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of MnBr₂ = 429.4 g, M of MnBr₂ = 214.7 g/mol, M of HBr = 80.9 g/mol\nStep 1 — convert the target mass to moles:\nn = m / M\nn of MnBr₂ = 429.4 / 214.7\nn of MnBr₂ = 2.0 mol\nStep 2 — convert to moles of HBr with the mole ratio:\nmoles of HBr = moles of MnBr₂ × (coefficient of HBr ÷ coefficient of MnBr₂)\nmoles of HBr = 2.0 × (2 ÷ 1)\nmoles of HBr = 4.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of HBr = 4.0 × 80.9\nm of HBr = 323.6 g"
}
STO-008 · medium · application · seq 16 · fully authored (F3 reverse pairing; fresh combustion surface 2C5H10 + 15O2 -> 10CO2 + 10H2O; the withheld corpus-banned M(CO2)=44.0 appears ONLY inside the wrong-M misconception feedback, per the STO-008 specNotes name-the-withheld-value convention) · numeric
In 2C₅H₁₀ + 15O₂ → 10CO₂ + 10H₂O, what mass of C₅H₁₀ must burn completely to make 7.0 mol of CO₂? (M of C₅H₁₀ = 70.0 g/mol) Give your answer in grams to one decimal place.
Answer: 98.0 g KEY
also accepted: 98.0 g
wrongValueDiagnostics
[
{
"value": 490.0,
"misconception": "you skipped the mole-ratio step — 7.0 mol is the CO₂'s moles, and only a fifth as many moles of C₅H₁₀ are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 7.0 mol is the CO₂'s moles, and only a fifth as many moles of C₅H₁₀ are needed."
},
{
"value": 2450.0,
"misconception": "you turned the fraction upside down — the wanted substance C₅H₁₀'s coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance C₅H₁₀'s coefficient 2 sits on top."
},
{
"value": 61.6,
"misconception": "you used CO₂'s molar mass (44.0 g/mol) — step 2 needs the WANTED substance C₅H₁₀'s molar mass",
"feedback": "This answer is wrong — you used CO₂'s molar mass (44.0 g/mol) — step 2 needs the WANTED substance C₅H₁₀'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of CO₂ = 7.0 mol, M of C₅H₁₀ = 70.0 g/mol\nStep 1 — convert to moles of C₅H₁₀ with the mole ratio:\nmoles of C₅H₁₀ = moles of CO₂ × (coefficient of C₅H₁₀ ÷ coefficient of CO₂)\nmoles of C₅H₁₀ = 7.0 × (2 ÷ 10)\nmoles of C₅H₁₀ = 1.4 mol\nStep 2 — convert the moles to mass:\nm = n × M\nm of C₅H₁₀ = 1.4 × 70.0\nm of C₅H₁₀ = 98.0 g"
}
STO-011 · hard · transfer · seq 17 · authored (fresh audit per blueprint case rotation; error class: product coefficient ignored) · mcq4
For 2Li + F₂ → 2LiF, a student totals the two sides and gets 51.8 g of reactants but only 25.9 g of products. (M of Li = 6.9 g/mol, M of F₂ = 38.0 g/mol, M of LiF = 25.9 g/mol.) What went wrong?
The coefficient 2 on LiF was ignored — the product side is 2 × 25.9 = 51.8 g. KEY
The reactant total is wrong — the F₂ needs a coefficient greater than 1 to balance the fluorine.
The equation itself is unbalanced, so the totals cannot match.
Nothing went wrong — some of the F₂ escapes as a gas, so the products weigh less.
STO-013 · hard · transfer · seq 18 · authored to seed:U8-G1.pool[38] class (STO-013 gateItems[2], step-2 flipped-ratio) — instance freshened (N2 + O2 → 2NO is STO-002 WE0 corpus; 28.0/30.0 banned) · mcq4
For 2Ag + S → Ag₂S, a student calculates the mass of Ag₂S formed from 431.6 g of Ag. (M of Ag = 107.9 g/mol, M of Ag₂S = 247.9 g/mol.)
Step 1: n of Ag = 431.6 / 107.9 = 4.0 mol
Step 2: moles of Ag₂S = 4.0 × 2/1 = 8.0 mol
Step 3: m of Ag₂S = 8.0 × 247.9 = 1983.2 g
Which step contains the error?
Step 2 KEY
Step 1
Step 3
No step contains an error
STO-010 · hard · transfer · seq 19 · fully authored per blueprint (production-planning framing on an unfamiliar precipitation surface, awkward 2:1 pair, both Ms supplied; wrongs: skipped ratio / inverted / wrong-M) · numeric
A kit supplier prepares golden-rain demonstration kits and must produce 414.9 g of PbI₂ in one batch. In Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃, what mass of KI must react completely to deliver that batch? (M of PbI₂ = 461.0 g/mol; M of KI = 166.0 g/mol) Give your answer in grams to one decimal place.
Answer: 298.8 g KEY
also accepted: 298.8 g
wrongValueDiagnostics
[
{
"value": 149.4,
"misconception": "you skipped the mole-ratio step — 0.90 mol is the PbI₂'s moles, and twice as many moles of KI are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 0.90 mol is the PbI₂'s moles, and twice as many moles of KI are needed."
},
{
"value": 74.7,
"misconception": "you turned the fraction upside down — the wanted substance KI's coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance KI's coefficient 2 sits on top."
},
{
"value": 829.8,
"misconception": "you used PbI₂'s molar mass in step 3 — the final step needs the WANTED substance KI's molar mass",
"feedback": "This answer is wrong — you used PbI₂'s molar mass in step 3 — the final step needs the WANTED substance KI's molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of PbI₂ = 414.9 g, M of PbI₂ = 461.0 g/mol, M of KI = 166.0 g/mol\nStep 1 — convert the target mass to moles:\nn = m / M\nn of PbI₂ = 414.9 / 461.0\nn of PbI₂ = 0.90 mol\nStep 2 — convert to moles of KI with the mole ratio:\nmoles of KI = moles of PbI₂ × (coefficient of KI ÷ coefficient of PbI₂)\nmoles of KI = 0.90 × (2 ÷ 1)\nmoles of KI = 1.8 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of KI = 1.8 × 166.0\nm of KI = 298.8 g"
}
STO-009 · hard · transfer · seq 20 · fully authored per blueprint (capstone terminal slot: full forward chain, awkward 3:1 pair, mixed-oxide transfer surface; reactant→product only) · numeric
In 3Mn + 2O₂ → Mn₃O₄, 494.1 g of Mn powder (M = 54.9 g/mol) burns completely. What mass of Mn₃O₄ forms? (M of Mn₃O₄ = 228.7 g/mol) Give your answer in grams to one decimal place.
Answer: 686.1 g KEY
also accepted: 686.1 g
wrongValueDiagnostics
[
{
"value": 2058.3,
"misconception": "you skipped the mole-ratio step — 9.0 mol is the Mn's moles, and only a third as many moles of Mn₃O₄ form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 9.0 mol is the Mn's moles, and only a third as many moles of Mn₃O₄ form."
},
{
"value": 6174.9,
"misconception": "you turned the fraction upside down — the wanted substance Mn₃O₄'s coefficient 1 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Mn₃O₄'s coefficient 1 sits on top."
},
{
"value": 164.7,
"misconception": "you used Mn's molar mass in step 3 — the final step needs the WANTED substance Mn₃O₄'s molar mass",
"feedback": "This answer is wrong — you used Mn's molar mass in step 3 — the final step needs the WANTED substance Mn₃O₄'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Mn = 494.1 g, M of Mn = 54.9 g/mol, M of Mn₃O₄ = 228.7 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Mn = 494.1 / 54.9\nn of Mn = 9.0 mol\nStep 2 — convert to moles of Mn₃O₄ with the mole ratio:\nmoles of Mn₃O₄ = moles of Mn × (coefficient of Mn₃O₄ ÷ coefficient of Mn)\nmoles of Mn₃O₄ = 9.0 × (1 ÷ 3)\nmoles of Mn₃O₄ = 3.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of Mn₃O₄ = 3.0 × 228.7\nm of Mn₃O₄ = 686.1 g"
}
STO-005 · medium · application · seq 21 · authored (FRQ1 per composition.frq_split_rationale: STO-005 mole-to-mole in BOTH roles + taught sanity check; mole-only, no molar masses per the STO-007+ boundary; fresh reaction + numbers per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
Ca reacts with H₂O according to the balanced equation Ca + 2H₂O → Ca(OH)2 + H₂. Work in moles throughout — no molar masses are needed.
Ca reacts with H₂O according to the balanced equation Ca + 2H₂O → Ca(OH)2 + H₂. Work in moles throughout — no molar masses are needed.
(a) 10.0 mol of H₂O reacts completely with Ca. Calculate the moles of H₂ that form. Show your working: state the given moles, write the mole-ratio equation with both substances named, substitute, and calculate.
(b) Calculate the moles of H₂O needed to make 4.0 mol of Ca(OH)2. Show the same working.
(c) For part (a), apply the lesson's sanity check without recalculating: compare the coefficient of the wanted substance with the coefficient of the given substance, state whether the answer should come out smaller or larger than the given moles, and confirm that your part (a) answer passes the check.
Part a. 10.0 mol of H₂O reacts completely with Ca. Calculate the moles of H₂ that form. Show your working: state the given moles, write the mole-ratio equation with both substances named, substitute, and calculate.
Model: moles of water = 10.0 mol
moles of hydrogen = moles of water × (coefficient of hydrogen ÷ coefficient of water)
moles of hydrogen = 10.0 × (1 ÷ 2)
moles of hydrogen = 5.0 mol
Part b. Calculate the moles of H₂O needed to make 4.0 mol of Ca(OH)2. Show the same working.
Model: moles of Ca(OH)2 = 4.0 mol (the target)
moles of water = moles of Ca(OH)2 × (coefficient of water ÷ coefficient of Ca(OH)2)
moles of water = 4.0 × (2 ÷ 1)
moles of water = 8.0 mol
Part c. For part (a), apply the lesson's sanity check without recalculating: compare the coefficient of the wanted substance with the coefficient of the given substance, state whether the answer should come out smaller or larger than the given moles, and confirm that your part (a) answer passes the check.
Model: The taught check says: when the wanted coefficient is smaller than the given one, the answer comes out smaller than the given moles. In part (a) the wanted substance hydrogen has coefficient 1 and the given substance water has coefficient 2 — the wanted coefficient is smaller, so the answer should come out smaller than 10.0 mol. The answer 5.0 mol is smaller than 10.0 mol, so it passes the check.
Full model answer. (a)
moles of water = 10.0 mol
moles of hydrogen = moles of water × (coefficient of hydrogen ÷ coefficient of water)
moles of hydrogen = 10.0 × (1 ÷ 2)
moles of hydrogen = 5.0 mol
(b)
moles of Ca(OH)2 = 4.0 mol (the target)
moles of water = moles of Ca(OH)2 × (coefficient of water ÷ coefficient of Ca(OH)2)
moles of water = 4.0 × (2 ÷ 1)
moles of water = 8.0 mol
(c)
The taught check says: when the wanted coefficient is smaller than the given one, the answer comes out smaller than the given moles. In part (a) the wanted substance hydrogen has coefficient 1 and the given substance water has coefficient 2 — the wanted coefficient is smaller, so the answer should come out smaller than 10.0 mol. The answer 5.0 mol is smaller than 10.0 mol, so it passes the check.
STO-009 · hard · application · seq 22 · authored (FRQ2 per composition.frq_split_rationale: STO-009 capstone chain end-to-end — plan + full mass-to-mass run + step-3 molar-mass rule; forward reactant→product only per the STO-010 boundary; both Ms supplied, corpus-free (Li from the blueprint's fresh-metal palette; Li uniquely forms the normal oxide, so 4Li + O2 → 2Li2O is chemically exact); fresh reaction + masses per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
Li burns in O₂ to form Li₂O as the balanced equation shows 4Li + O₂ → 2Li₂O. 55.2 g of Li reacts completely. (M of Li = 6.9 g/mol; M of Li₂O = 29.8 g/mol.)
Li burns in O₂ to form Li₂O as the balanced equation shows 4Li + O₂ → 2Li₂O. 55.2 g of Li reacts completely. (M of Li = 6.9 g/mol; M of Li₂O = 29.8 g/mol.)
(a) Before calculating, state the three-step plan for working out the mass of Li₂O that forms, naming the tool each step uses.
(b) Carry out the calculation, showing the working for every step, and give the mass of Li₂O that forms.
(c) State which substance's molar mass belongs in step 3, and why.
Part a. Before calculating, state the three-step plan for working out the mass of Li₂O that forms, naming the tool each step uses.
Model: grams of Li → moles of Li (n = m/M) → moles of the lithium oxide (mole-ratio equation) → grams of the lithium oxide (m = n × M).
Part b. Carry out the calculation, showing the working for every step, and give the mass of Li₂O that forms.
Model: m of Li = 55.2 g; M of Li = 6.9 g/mol; M of the lithium oxide = 29.8 g/mol
Step 1 — convert the mass to moles:
n = m / M
n of Li = 55.2 / 6.9
n of Li = 8.0 mol
Step 2 — convert to moles of the lithium oxide with the mole ratio:
moles of the lithium oxide = moles of Li × (coefficient of the lithium oxide ÷ coefficient of Li)
moles of the lithium oxide = 8.0 × (2 ÷ 4)
moles of the lithium oxide = 4.0 mol
Step 3 — convert the moles to mass:
m = n × M
m of the lithium oxide = 4.0 × 29.8
m of the lithium oxide = 119.2 g
Part c. State which substance's molar mass belongs in step 3, and why.
Model: Step 3 must use the WANTED substance's molar mass — the lithium oxide's, 29.8 g/mol — because step 3 converts the wanted substance's moles into its mass with m = n × M. The given substance's molar mass (Li's 6.9 g/mol) is used only in step 1 and never appears in step 3.
Full model answer. (a)
grams of Li → moles of Li (n = m/M) → moles of the lithium oxide (mole-ratio equation) → grams of the lithium oxide (m = n × M).
(b)
m of Li = 55.2 g; M of Li = 6.9 g/mol; M of the lithium oxide = 29.8 g/mol
Step 1 — convert the mass to moles:
n = m / M
n of Li = 55.2 / 6.9
n of Li = 8.0 mol
Step 2 — convert to moles of the lithium oxide with the mole ratio:
moles of the lithium oxide = moles of Li × (coefficient of the lithium oxide ÷ coefficient of Li)
moles of the lithium oxide = 8.0 × (2 ÷ 4)
moles of the lithium oxide = 4.0 mol
Step 3 — convert the moles to mass:
m = n × M
m of the lithium oxide = 4.0 × 29.8
m of the lithium oxide = 119.2 g
(c)
Step 3 must use the WANTED substance's molar mass — the lithium oxide's, 29.8 g/mol — because step 3 converts the wanted substance's moles into its mass with m = n × M. The given substance's molar mass (Li's 6.9 g/mol) is used only in step 1 and never appears in step 3.
STO-013 · hard · transfer · seq 23 · authored (FRQ3 per composition.frq_split_rationale: STO-013 inspection PRODUCTION — one planted error; C-form class = step-1 wrong-M, self-consistent downstream (atomic mass 1.0 used instead of M of H2 = 2.0; both values supplied in-stem); fresh reaction + numbers per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
For H₂ + Cl₂ → 2HCl, a student calculates the mass of HCl formed when 6.0 g of H₂ reacts completely. (Atomic masses: H 1.0, Cl 35.5; M of H₂ = 2.0 g/mol; M of HCl = 36.5 g/mol.)
Step 1: n of H₂ = 6.0 / 1.0 = 6.0 mol
Step 2: moles of HCl = 6.0 × (2 ÷ 1) = 12.0 mol
Step 3: m of HCl = 12.0 × 36.5 = 438.0 g
Exactly one step contains an error.
For H₂ + Cl₂ → 2HCl, a student calculates the mass of HCl formed when 6.0 g of H₂ reacts completely. (Atomic masses: H 1.0, Cl 35.5; M of H₂ = 2.0 g/mol; M of HCl = 36.5 g/mol.)
Step 1: n of H₂ = 6.0 / 1.0 = 6.0 mol
Step 2: moles of HCl = 6.0 × (2 ÷ 1) = 12.0 mol
Step 3: m of HCl = 12.0 × 36.5 = 438.0 g
Exactly one step contains an error.
(a) Name the step that contains the error, and state what is wrong with it.
(b) Correct the erroneous step, rerun the working from that point, and give the corrected mass of HCl.
(c) A classmate claims steps 2 and 3 must also contain errors, because every number after step 1 came out wrong. Using the lesson's rule about steps that inherit a wrong number, state whether the classmate is right, and explain.
Part a. Name the step that contains the error, and state what is wrong with it.
Model: The error is in step 1. Step 1 must divide the given mass by the given substance's own molar mass: n = m / M with M of hydrogen = 2.0 g/mol. The student divided by 1.0 — the atomic mass of H — instead of the molar mass of hydrogen. Steps 2 and 3 use correct methods, so the wrong molar mass in step 1 is the one planted error.
Part b. Correct the erroneous step, rerun the working from that point, and give the corrected mass of HCl.
Model: Corrected step 1:
m of hydrogen = 6.0 g; M of hydrogen = 2.0 g/mol
n = m / M
n of hydrogen = 6.0 / 2.0
n of hydrogen = 3.0 mol
Rerun step 2:
moles of HCl = moles of hydrogen × (coefficient of HCl ÷ coefficient of hydrogen)
moles of HCl = 3.0 × (2 ÷ 1)
moles of HCl = 6.0 mol
Rerun step 3:
m = n × M
m of HCl = 6.0 × 36.5
m of HCl = 219.0 g
Part c. A classmate claims steps 2 and 3 must also contain errors, because every number after step 1 came out wrong. Using the lesson's rule about steps that inherit a wrong number, state whether the classmate is right, and explain.
Model: Steps 2 and 3 do not contain errors. The taught rule: a step that inherits a wrong number from an earlier step is not itself the error — the error lives where the wrong number was made. Steps 2 and 3 used correct methods on the moles handed to them; they inherited step 1's wrong 6.0 mol and carried it faithfully to the end, which is why the final mass is wrong even though only step 1 is in error.
Full model answer. (a)
The error is in step 1. Step 1 must divide the given mass by the given substance's own molar mass: n = m / M with M of hydrogen = 2.0 g/mol. The student divided by 1.0 — the atomic mass of H — instead of the molar mass of hydrogen. Steps 2 and 3 use correct methods, so the wrong molar mass in step 1 is the one planted error.
(b)
Corrected step 1:
m of hydrogen = 6.0 g; M of hydrogen = 2.0 g/mol
n = m / M
n of hydrogen = 6.0 / 2.0
n of hydrogen = 3.0 mol
Rerun step 2:
moles of HCl = moles of hydrogen × (coefficient of HCl ÷ coefficient of hydrogen)
moles of HCl = 3.0 × (2 ÷ 1)
moles of HCl = 6.0 mol
Rerun step 3:
m = n × M
m of HCl = 6.0 × 36.5
m of HCl = 219.0 g
(c)
Steps 2 and 3 do not contain errors. The taught rule: a step that inherits a wrong number from an earlier step is not itself the error — the error lives where the wrong number was made. Steps 2 and 3 used correct methods on the moles handed to them; they inherited step 1's wrong 6.0 mol and carried it faithfully to the end, which is why the final mass is wrong even though only step 1 is in error.
The balanced equation 2K + Br₂ → 2KBr can be read in more than one counting unit. Which option gives a correct reading AND the correct reason it holds?
2 mol of K react with 1 mol of Br₂ — because a mole is the same fixed count for every substance, so scaling every substance up by that same count leaves the ratio unchanged. KEY
2 g of K react with 1 g of Br₂ — because the coefficients fix the reacting masses directly, so exactly the same two numbers carry straight over into the gram amounts.
2 mol of K react with 1 mol of Br₂ — because K and Br₂ have equal molar masses, so equal counts always pair up.
2 mol of K react with 1 mol of Br₂ — because the mole is defined separately for each reaction so that its coefficients come out right.
STO-001 · easy · recall · seq 2 · authored (fresh burn with both coefficients written, coprime as asked; fresh instance 2CH₃OH + 3O₂ → 2CO₂ + 4H₂O) · text
In the balanced equation 2CH₃OH + 3O₂ → 2CO₂ + 4H₂O, what is the mole ratio of O₂ to H₂O? Write the ratio with a colon, such as 4:1.
Answer: 3:4 KEY
also accepted: 3 : 4, 3 to 4
wrongValueDiagnostics
[
{
"value": "4:3",
"misconception": "you reversed the order — the question asks O₂ to H₂O, so O₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order. The question asks O₂ to H₂O, so O₂'s coefficient 3 comes first: 3:4."
},
{
"value": "3:2",
"misconception": "you took CO₂'s coefficient instead of H₂O's — the second number must belong to the substance named second",
"feedback": "This answer is wrong — you took CO₂'s coefficient 2 instead of H₂O's. The second number must belong to the substance named second, H₂O, whose coefficient is 4: 3:4."
},
{
"value": "1:1",
"misconception": "you ignored the coefficients — the mole ratio is read from the written coefficients, 3 for O₂ and 4 for H₂O",
"feedback": "This answer is wrong — you ignored the coefficients. The mole ratio comes from the written coefficients: 3 for O₂ and 4 for H₂O, so the ratio is 3:4."
}
]
feedback
{
"keyExplanation": "The coefficient of O₂ is 3, and the coefficient of H₂O is 4. Take the coefficients in the order asked: O₂ first, H₂O second. The mole ratio of O₂ to H₂O is 3:4.",
"anticipatedWrong": [
{
"value": "4:3",
"misconception": "you reversed the order — the question asks O₂ to H₂O, so O₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order. The question asks O₂ to H₂O, so O₂'s coefficient 3 comes first: 3:4."
},
{
"value": "3:2",
"misconception": "you took CO₂'s coefficient instead of H₂O's — the second number must belong to the substance named second",
"feedback": "This answer is wrong — you took CO₂'s coefficient 2 instead of H₂O's. The second number must belong to the substance named second, H₂O, whose coefficient is 4: 3:4."
},
{
"value": "1:1",
"misconception": "you ignored the coefficients — the mole ratio is read from the written coefficients, 3 for O₂ and 4 for H₂O",
"feedback": "This answer is wrong — you ignored the coefficients. The mole ratio comes from the written coefficients: 3 for O₂ and 4 for H₂O, so the ratio is 3:4."
}
]
}
A student needs the mole ratio of H₂ to HBr. The unbalanced equation H₂ + Br₂ → HBr suggests one ratio, and the balanced equation H₂ + Br₂ → 2HBr suggests another. Why is only the balanced version trustworthy?
Only the balanced coefficients count atoms correctly — the unbalanced version shows atoms disappearing, and a reaction only rearranges atoms. KEY
The unbalanced version is trustworthy too, as long as only a small amount of H₂ reacts.
The unbalanced version runs the reaction backwards, so its ratio describes HBr breaking apart instead of forming.
Every unbalanced equation understates each coefficient by exactly half, so doubling every number makes it trustworthy.
STO-006 · easy · recall · seq 4 · authored (route-anatomy rotation per blueprint — where-the-molar-masses-enter read, instance-free, zero arithmetic per JM-P1a) · mcq4
A mass-to-mass calculation uses two molar masses. Where do they enter the route?
In the first and last steps — each one converts a single substance between its own mass and its own moles. KEY
In the middle step — the mole ratio divides one molar mass by the other.
In every step — all three steps need both molar masses to keep the units consistent.
In the last step only — both molar masses multiply the final moles together.
STO-004 · medium · application · seq 5 · authored (parallel mcq slot; fresh forward setup on a metal-halide synthesis; divide-by-fraction distractor class per STO-004 specNotes) · mcq4
For 2Cr + 3Cl₂ → 2CrCl₃, which equation finds the moles of CrCl₃ formed from the moles of Cl₂?
In 2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O, 1.0 mol of C₃H₇OH burns completely. How many moles of H₂O form? Give your answer in moles to one decimal place.
Answer: 4.0 mol KEY
also accepted: 4.0 mol
wrongValueDiagnostics
[
{
"value": 0.25,
"misconception": "you turned the fraction upside down — the wanted substance H₂O's coefficient 8 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance H₂O's coefficient 8 sits on top."
},
{
"value": 1.0,
"misconception": "you ignored the coefficients and matched the moles one-for-one — the ratio of H₂O to C₃H₇OH is 8:2",
"feedback": "This answer is wrong — you ignored the coefficients and matched the moles one-for-one — the ratio of H₂O to C₃H₇OH is 8:2."
},
{
"value": 8.0,
"misconception": "you multiplied by H₂O's coefficient alone instead of the coefficient ratio",
"feedback": "This answer is wrong — you multiplied by H₂O's coefficient alone instead of the coefficient ratio."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of C₃H₇OH = 1.0 mol\nWrite down the equation:\nmoles of H₂O = moles of C₃H₇OH × (coefficient of H₂O ÷ coefficient of C₃H₇OH)\nSubstitute in the values, and calculate:\nmoles of H₂O = 1.0 × (8 ÷ 2)\nmoles of H₂O = 4.0 mol"
}
STO-001 · medium · application · seq 7 · authored (parallel text slot; fresh unwritten-1 pair with the inside-subscript trap on the asked substance P₄) · text
In the balanced equation P₄ + 5O₂ → P₄O₁₀, what is the mole ratio of P₄ to O₂? Write the ratio with a colon, such as 4:1.
Answer: 1:5 KEY
also accepted: 1 : 5, 1 to 5
wrongValueDiagnostics
[
{
"value": "4:5",
"misconception": "you used the subscript 4 inside P₄ as its coefficient — P₄ has no written coefficient, so its coefficient is 1",
"feedback": "This answer is wrong — you used the subscript 4 inside P₄ as its coefficient — P₄ has no written coefficient, so its coefficient is 1."
},
{
"value": "5:1",
"misconception": "you reversed the order — the question asks P₄ to O₂, so P₄'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order — the question asks P₄ to O₂, so P₄'s coefficient comes first."
},
{
"value": "1:1",
"misconception": "you took P₄O₁₀'s coefficient for O₂ — the second number must belong to O₂, whose coefficient is 5",
"feedback": "This answer is wrong — you took P₄O₁₀'s coefficient for O₂ — the second number must belong to O₂, whose coefficient is 5."
}
]
feedback
{
"keyExplanation": "P₄ has no written coefficient, so its coefficient is 1; the coefficient of O₂ is 5. The subscript 4 inside P₄ counts atoms in one molecule and plays no part in the mole ratio. The mole ratio of P₄ to O₂ is 1:5.",
"anticipatedWrong": [
{
"value": "4:5",
"misconception": "you used the subscript 4 inside P₄ as its coefficient — P₄ has no written coefficient, so its coefficient is 1",
"feedback": "This answer is wrong — you used the subscript 4 inside P₄ as its coefficient — P₄ has no written coefficient, so its coefficient is 1."
},
{
"value": "5:1",
"misconception": "you reversed the order — the question asks P₄ to O₂, so P₄'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order — the question asks P₄ to O₂, so P₄'s coefficient comes first."
},
{
"value": "1:1",
"misconception": "you took P₄O₁₀'s coefficient for O₂ — the second number must belong to O₂, whose coefficient is 5",
"feedback": "This answer is wrong — you took P₄O₁₀'s coefficient for O₂ — the second number must belong to O₂, whose coefficient is 5."
}
]
}
STO-007 · medium · application · seq 8 · authored (parallel numeric slot; fresh metal + halogen surface per the case rotation; only the given substance's M supplied) · numeric
In Sn + 2Br₂ → SnBr₄, 237.4 g of Sn (M = 118.7 g/mol) reacts completely with Br₂. How many moles of Br₂ are used up? Give your answer in moles to 2 significant figures.
Answer: 4.0 mol KEY
also accepted: 4.0 mol
wrongValueDiagnostics
[
{
"value": 1.0,
"misconception": "you turned the fraction upside down — the wanted substance Br₂'s coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Br₂'s coefficient 2 sits on top."
},
{
"value": 2.0,
"misconception": "you skipped the mole-ratio step — 2.0 mol is the Sn's moles, not the Br₂'s",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 2.0 mol is the Sn's moles, not the Br₂'s."
},
{
"value": 474.8,
"misconception": "you applied the mole ratio to the mass instead of converting to moles first",
"feedback": "This answer is wrong — you applied the mole ratio to the mass instead of converting to moles first."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Sn = 237.4 g, M of Sn = 118.7 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Sn = 237.4 / 118.7\nn of Sn = 2.0 mol\nStep 2 — convert to moles of Br₂ with the mole ratio:\nmoles of Br₂ = moles of Sn × (coefficient of Br₂ ÷ coefficient of Sn)\nmoles of Br₂ = 2.0 × (2 ÷ 1)\nmoles of Br₂ = 4.0 mol"
}
STO-008 · medium · application · seq 9 · authored (parallel numeric slot; fresh halide-synthesis instance with an unwritten-1 wanted coefficient) · numeric
In Si + 2Cl₂ → SiCl₄, 6.0 mol of Cl₂ reacts completely with Si. What mass of SiCl₄ forms? (M of SiCl₄ = 170.1 g/mol) Give your answer in grams to one decimal place.
Answer: 510.3 g KEY
also accepted: 510.3 g
wrongValueDiagnostics
[
{
"value": 1020.6,
"misconception": "you skipped the mole-ratio step — 6.0 mol is the Cl₂'s moles, and only half as many moles of SiCl₄ form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 6.0 mol is the Cl₂'s moles, and only half as many moles of SiCl₄ form."
},
{
"value": 2041.2,
"misconception": "you turned the fraction upside down — the wanted substance SiCl₄'s coefficient 1 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance SiCl₄'s coefficient 1 sits on top."
},
{
"value": 213.0,
"misconception": "you used Cl₂'s molar mass (71.0 g/mol) — step 2 needs the WANTED substance SiCl₄'s molar mass",
"feedback": "This answer is wrong — you used Cl₂'s molar mass (71.0 g/mol) — step 2 needs the WANTED substance SiCl₄'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of Cl₂ = 6.0 mol, M of SiCl₄ = 170.1 g/mol\nStep 1 — convert to moles of SiCl₄ with the mole ratio:\nmoles of SiCl₄ = moles of Cl₂ × (coefficient of SiCl₄ ÷ coefficient of Cl₂)\nmoles of SiCl₄ = 6.0 × (1 ÷ 2)\nmoles of SiCl₄ = 3.0 mol\nStep 2 — convert the moles to mass:\nm = n × M\nm of SiCl₄ = 3.0 × 170.1\nm of SiCl₄ = 510.3 g"
}
STO-012 · medium · application · seq 10 · authored (defense-class rotation on a fresh metal — lead, the one blueprint-listed metal with no equation in the authored bank; 828.8 + 64.0 = 892.8 exact, all three numbers corpus-free and bank-fresh) · mcq4
A student heats 828.8 g of lead in a stream of O₂ until it has reacted completely to form 892.8 g of lead(II) oxide, PbO, and says the result breaks the law of conservation of mass. Which reply is correct?
It does not — the extra 64.0 g is the O₂ that reacted, and counting both reactants the mass balances exactly. KEY
It does — the law of conservation of mass only holds for reactions run inside sealed containers.
It does not — the law allows small gains when a reaction gives out light and heat.
It does not — lead(II) oxide is simply a denser substance, and denser substances weigh more.
STO-005 · medium · application · seq 11 · authored (fresh O2-needed burn instance on a lab-solvent fuel spoken by formula per the symbol convention) · numeric
In C₃H₆O + 4O₂ → 3CO₂ + 3H₂O, how many moles of O₂ are needed to burn 3.0 mol of C₃H₆O completely? Give your answer in moles to one decimal place.
Answer: 12.0 mol KEY
also accepted: 12.0 mol
wrongValueDiagnostics
[
{
"value": 0.8,
"misconception": "you turned the fraction upside down — the wanted substance O₂'s coefficient 4 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance O₂'s coefficient 4 sits on top."
},
{
"value": 3.0,
"misconception": "you ignored the coefficients and matched the moles one-for-one — the ratio of O₂ to C₃H₆O is 4:1",
"feedback": "This answer is wrong — you ignored the coefficients and matched the moles one-for-one — the ratio of O₂ to C₃H₆O is 4:1."
},
{
"value": 9.0,
"misconception": "you used CO₂'s coefficient — the wanted substance is O₂, whose coefficient is 4",
"feedback": "This answer is wrong — you used CO₂'s coefficient — the wanted substance is O₂, whose coefficient is 4."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of C₃H₆O = 3.0 mol\nWrite down the equation:\nmoles of O₂ = moles of C₃H₆O × (coefficient of O₂ ÷ coefficient of C₃H₆O)\nSubstitute in the values, and calculate:\nmoles of O₂ = 3.0 × (4 ÷ 1)\nmoles of O₂ = 12.0 mol"
}
STO-009 · medium · application · seq 12 · authored (parallel numeric slot; fresh 2:4 metal-oxide chain — 4Mn+3O2->2Mn2O3 is bank-fresh, existing rows use only 3Mn+2O2->Mn3O4; O2's M never supplied by construction) · numeric
In 4Mn + 3O₂ → 2Mn₂O₃, 219.6 g of Mn (M = 54.9 g/mol) reacts completely. (M of Mn₂O₃ = 157.8 g/mol) How many grams of Mn₂O₃ form? Give your answer to one decimal place.
Answer: 315.6 g KEY
also accepted: 315.6 g
wrongValueDiagnostics
[
{
"value": 631.2,
"misconception": "you skipped the mole-ratio step — 4.0 mol is the Mn's moles, and only half as many moles of Mn₂O₃ form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 4.0 mol is the Mn's moles, and only half as many moles of Mn₂O₃ form."
},
{
"value": 1262.4,
"misconception": "you turned the fraction upside down — the wanted substance Mn₂O₃'s coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Mn₂O₃'s coefficient 2 sits on top."
},
{
"value": 109.8,
"misconception": "you used Mn's molar mass in step 3 — the final step needs the WANTED substance Mn₂O₃'s molar mass",
"feedback": "This answer is wrong — you used Mn's molar mass in step 3 — the final step needs the WANTED substance Mn₂O₃'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Mn = 219.6 g, M of Mn = 54.9 g/mol, M of Mn₂O₃ = 157.8 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Mn = 219.6 / 54.9\nn of Mn = 4.0 mol\nStep 2 — convert to moles of Mn₂O₃ with the mole ratio:\nmoles of Mn₂O₃ = moles of Mn × (coefficient of Mn₂O₃ ÷ coefficient of Mn)\nmoles of Mn₂O₃ = 4.0 × (2 ÷ 4)\nmoles of Mn₂O₃ = 2.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of Mn₂O₃ = 2.0 × 157.8\nm of Mn₂O₃ = 315.6 g"
}
STO-011 · medium · application · seq 13 · authored (parallel numeric slot; fresh instance 2Bi + 3Cl2 -> 2BiCl3; M of Cl2 = 71.0 is corpus-banned so only the product side's M is supplied and asked) · numeric
Bi reacts completely with Cl₂: 2Bi + 3Cl₂ → 2BiCl₃. (M of BiCl₃ = 315.5 g/mol.) What total mass of products does the balanced equation represent? Give your answer in grams to one decimal place.
Answer: 631.0 g KEY
also accepted: 631.0 g
wrongValueDiagnostics
[
{
"value": 315.5,
"misconception": "you ignored the coefficient 2 on BiCl₃ — its mass is 2 × 315.5 g",
"feedback": "This answer is wrong — you ignored the coefficient 2 on BiCl₃ — its mass is 2 × 315.5 g."
},
{
"value": 946.5,
"misconception": "you multiplied by 3, the Cl₂ coefficient, instead of by BiCl₃'s own coefficient 2",
"feedback": "This answer is wrong — you multiplied by 3, the Cl₂ coefficient, instead of by BiCl₃'s own coefficient 2."
},
{
"value": 317.5,
"misconception": "you added the coefficient instead of multiplying by it",
"feedback": "This answer is wrong — you added the coefficient instead of multiplying by it."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM of BiCl₃ = 315.5 g/mol, coefficient of BiCl₃ = 2\nWrite down the equation:\nmass of each substance = coefficient × molar mass\nSubstitute in the values, and calculate:\ntotal mass of products = 2 × 315.5\ntotal mass of products = 631.0 g"
}
STO-007 · medium · application · seq 14 · authored (parallel numeric slot; PRODUCT-mass-given read across the arrow; wave-10 sol realism fix: 2Co+O2->2CoO dropped (Co oxidation favors Co3O4) => direct nitridation 2Al + N2 -> 2AlN (industrial AlN route); ZnO and CaO repicks census-killed; value set 13.0/26.0/6.5 retained) · numeric
Al is heated in N₂ and reacts completely: 2Al + N₂ → 2AlN. A batch forms 533.0 g of AlN (M = 41.0 g/mol). How many moles of N₂ were used up? Give your answer in moles to one decimal place.
Answer: 6.5 mol KEY
also accepted: 6.5 mol
wrongValueDiagnostics
[
{
"value": 26.0,
"misconception": "you turned the fraction upside down — the wanted substance N₂'s coefficient 1 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance N₂'s coefficient 1 sits on top."
},
{
"value": 13.0,
"misconception": "you stopped at step 1 — 13.0 mol is the AlN's moles, not the N₂'s",
"feedback": "This answer is wrong — you stopped at step 1 — 13.0 mol is the AlN's moles, not the N₂'s."
},
{
"value": 266.5,
"misconception": "you applied the mole ratio to the mass instead of converting to moles first",
"feedback": "This answer is wrong — you applied the mole ratio to the mass instead of converting to moles first."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of AlN = 533.0 g, M of AlN = 41.0 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of AlN = 533.0 / 41.0\nn of AlN = 13.0 mol\nStep 2 — convert to moles of N₂ with the mole ratio:\nmoles of N₂ = moles of AlN × (coefficient of N₂ ÷ coefficient of AlN)\nmoles of N₂ = 13.0 × (1 ÷ 2)\nmoles of N₂ = 6.5 mol"
}
STO-010 · medium · application · seq 15 · authored (parallel numeric slot; fresh metal-oxide reverse chain 4Bi + 3O2 -> 2Bi2O3, 4:2 ratio; O2 = forced-family lexicon token, its corpus-banned M never supplied) · numeric
In 4Bi + 3O₂ → 2Bi₂O₃, what mass of Bi must react completely with O₂ to make 349.5 g of Bi₂O₃? (M of Bi₂O₃ = 466.0 g/mol; M of Bi = 209.0 g/mol) Give your answer in grams to one decimal place.
Answer: 313.5 g KEY
also accepted: 313.5 g
wrongValueDiagnostics
[
{
"value": 156.8,
"misconception": "you skipped the mole-ratio step — 0.75 mol is the Bi₂O₃'s moles, and twice as many moles of Bi are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 0.75 mol is the Bi₂O₃'s moles, and twice as many moles of Bi are needed."
},
{
"value": 78.4,
"misconception": "you turned the fraction upside down — the wanted substance Bi's coefficient 4 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Bi's coefficient 4 sits on top."
},
{
"value": 699.0,
"misconception": "you used Bi₂O₃'s molar mass in step 3 — the final step needs the WANTED substance Bi's molar mass",
"feedback": "This answer is wrong — you used Bi₂O₃'s molar mass in step 3 — the final step needs the WANTED substance Bi's molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Bi₂O₃ = 349.5 g, M of Bi₂O₃ = 466.0 g/mol, M of Bi = 209.0 g/mol\nStep 1 — convert the target mass to moles:\nn = m / M\nn of Bi₂O₃ = 349.5 / 466.0\nn of Bi₂O₃ = 0.75 mol\nStep 2 — convert to moles of Bi with the mole ratio:\nmoles of Bi = moles of Bi₂O₃ × (coefficient of Bi ÷ coefficient of Bi₂O₃)\nmoles of Bi = 0.75 × (4 ÷ 2)\nmoles of Bi = 1.5 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of Bi = 1.5 × 209.0\nm of Bi = 313.5 g"
}
STO-008 · medium · application · seq 16 · fully authored (F3 reverse pairing with a 3:2 awkward ratio; fresh instance 2V + 3Br2 -> 2VBr3, distinct from the seq-13/15 vanadium-oxide instances; withheld M(VBr3)=290.6 named in the wrong-M misconception) · numeric
In 2V + 3Br₂ → 2VBr₃, what mass of Br₂ (M of Br₂ = 159.8 g/mol) is needed to react completely with vanadium to make 8.0 mol of VBr₃? Give your answer in grams to one decimal place.
Answer: 1917.6 g KEY
also accepted: 1917.6 g
wrongValueDiagnostics
[
{
"value": 1278.4,
"misconception": "you skipped the mole-ratio step — 8.0 mol is the VBr₃'s moles, and more moles of Br₂ are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 8.0 mol is the VBr₃'s moles, and more moles of Br₂ are needed."
},
{
"value": 852.3,
"misconception": "you turned the fraction upside down — the wanted substance Br₂'s coefficient 3 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Br₂'s coefficient 3 sits on top."
},
{
"value": 3487.2,
"misconception": "you used VBr₃'s molar mass (290.6 g/mol) — step 2 needs the WANTED substance Br₂'s molar mass",
"feedback": "This answer is wrong — you used VBr₃'s molar mass (290.6 g/mol) — step 2 needs the WANTED substance Br₂'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of VBr₃ = 8.0 mol, M of Br₂ = 159.8 g/mol\nStep 1 — convert to moles of Br₂ with the mole ratio:\nmoles of Br₂ = moles of VBr₃ × (coefficient of Br₂ ÷ coefficient of VBr₃)\nmoles of Br₂ = 8.0 × (3 ÷ 2)\nmoles of Br₂ = 12.0 mol\nStep 2 — convert the moles to mass:\nm = n × M\nm of Br₂ = 12.0 × 159.8\nm of Br₂ = 1917.6 g"
}
STO-011 · hard · transfer · seq 17 · authored (fresh audit per blueprint case rotation; error class: substance totaled on the wrong side) · mcq4
For Ni + 2HCl → NiCl₂ + H₂, a student totals the two sides and gets 133.7 g of reactants but only 129.7 g of products. (M of Ni = 58.7 g/mol, M of HCl = 36.5 g/mol, M of NiCl₂ = 129.7 g/mol, M of H₂ = 2.0 g/mol.) What went wrong?
The H₂ was totaled on the wrong side — it is a product, so the sides are 58.7 + 2 × 36.5 = 131.7 g of reactants and 129.7 + 2.0 = 131.7 g of products. KEY
The product total is wrong — NiCl₂ should be counted twice, because the two HCl together deliver two chloride units to the product.
The equation itself is unbalanced, so the totals cannot match.
Nothing went wrong — the H₂ gas escapes, so the products must weigh less than the reactants.
STO-013 · hard · transfer · seq 18 · authored (no-error case — the article mandates the class; key is the lesson's own taught 'No step contains an error' option, the strongest guard against it reading as filler) · mcq4
For 2Li + Br₂ → 2LiBr, a student calculates the mass of LiBr formed from 34.5 g of Li. (M of Li = 6.9 g/mol, M of LiBr = 86.8 g/mol.)
Step 1: n of Li = 34.5 / 6.9 = 5.0 mol
Step 2: moles of LiBr = 5.0 × 2/2 = 5.0 mol
Step 3: m of LiBr = 5.0 × 86.8 = 434.0 g
Which step contains the error?
No step contains an error KEY
Step 1
Step 2
Step 3
STO-010 · hard · transfer · seq 19 · fully authored per blueprint (production-planning framing on an unfamiliar industrial surface, awkward 3:1 pair, both Ms supplied; wrongs: skipped ratio / inverted / wrong-M) · numeric
In W + 3F₂ → WF₆, an etching-gas supplier must deliver 893.4 g of WF₆ for one order. What mass of F₂ must react completely with W to make it? (M of WF₆ = 297.8 g/mol; M of F₂ = 38.0 g/mol) Give your answer in grams to one decimal place.
Answer: 342.0 g KEY
also accepted: 342.0 g
wrongValueDiagnostics
[
{
"value": 114.0,
"misconception": "you skipped the mole-ratio step — 3.0 mol is the WF₆'s moles, and three times as many moles of F₂ are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 3.0 mol is the WF₆'s moles, and three times as many moles of F₂ are needed."
},
{
"value": 38.0,
"misconception": "you turned the fraction upside down — the wanted substance F₂'s coefficient 3 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance F₂'s coefficient 3 sits on top."
},
{
"value": 2680.2,
"misconception": "you used WF₆'s molar mass in step 3 — the final step needs the WANTED substance F₂'s molar mass",
"feedback": "This answer is wrong — you used WF₆'s molar mass in step 3 — the final step needs the WANTED substance F₂'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of WF₆ = 893.4 g, M of WF₆ = 297.8 g/mol, M of F₂ = 38.0 g/mol\nStep 1 — convert the target mass to moles:\nn = m / M\nn of WF₆ = 893.4 / 297.8\nn of WF₆ = 3.0 mol\nStep 2 — convert to moles of F₂ with the mole ratio:\nmoles of F₂ = moles of WF₆ × (coefficient of F₂ ÷ coefficient of WF₆)\nmoles of F₂ = 3.0 × (3 ÷ 1)\nmoles of F₂ = 9.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of F₂ = 9.0 × 38.0\nm of F₂ = 342.0 g"
}
STO-009 · hard · transfer · seq 20 · fully authored per blueprint (capstone terminal slot: full forward chain, awkward 2:1 pair, pigment transfer surface; reactant→product only) · numeric
A pigment plant fuses Cr with S: in 2Cr + 3S → Cr₂S₃, 416.0 g of Cr (M = 52.0 g/mol) reacts completely with S. What mass of Cr₂S₃ forms? (M of Cr₂S₃ = 200.3 g/mol) Give your answer in grams to one decimal place.
Answer: 801.2 g KEY
also accepted: 801.2 g
wrongValueDiagnostics
[
{
"value": 1602.4,
"misconception": "you skipped the mole-ratio step — 8.0 mol is the Cr's moles, and only half as many moles of Cr₂S₃ form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 8.0 mol is the Cr's moles, and only half as many moles of Cr₂S₃ form."
},
{
"value": 3204.8,
"misconception": "you turned the fraction upside down — the wanted substance Cr₂S₃'s coefficient 1 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Cr₂S₃'s coefficient 1 sits on top."
},
{
"value": 208.0,
"misconception": "you used Cr's molar mass in step 3 — the final step needs the WANTED substance Cr₂S₃'s molar mass",
"feedback": "This answer is wrong — you used Cr's molar mass in step 3 — the final step needs the WANTED substance Cr₂S₃'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Cr = 416.0 g, M of Cr = 52.0 g/mol, M of Cr₂S₃ = 200.3 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Cr = 416.0 / 52.0\nn of Cr = 8.0 mol\nStep 2 — convert to moles of Cr₂S₃ with the mole ratio:\nmoles of Cr₂S₃ = moles of Cr × (coefficient of Cr₂S₃ ÷ coefficient of Cr)\nmoles of Cr₂S₃ = 8.0 × (1 ÷ 2)\nmoles of Cr₂S₃ = 4.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of Cr₂S₃ = 4.0 × 200.3\nm of Cr₂S₃ = 801.2 g"
}
STO-005 · medium · application · seq 21 · authored (FRQ1 per composition.frq_split_rationale: STO-005 mole-to-mole in BOTH roles + taught sanity check; mole-only, no molar masses per the STO-007+ boundary; fresh reaction + numbers per form (butane fresh vs corpus fuels); equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
Butane, C₄H₁₀, burns according to the balanced equation 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O. Work in moles throughout — no molar masses are needed.
Butane, C₄H₁₀, burns according to the balanced equation 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O. Work in moles throughout — no molar masses are needed.
(a) 3.0 mol of C₄H₁₀ burns completely. Calculate the moles of CO₂ that form. Show your working: state the given moles, write the mole-ratio equation with both substances named, substitute, and calculate.
(b) Calculate the moles of C₄H₁₀ that must burn completely to make 20.0 mol of H₂O. Show the same working.
(c) For part (b), apply the lesson's sanity check without recalculating: compare the coefficient of the wanted substance with the coefficient of the given substance, state whether the answer should come out smaller or larger than the given moles, and confirm that your part (b) answer passes the check.
Part a. 3.0 mol of C₄H₁₀ burns completely. Calculate the moles of CO₂ that form. Show your working: state the given moles, write the mole-ratio equation with both substances named, substitute, and calculate.
Model: moles of butane = 3.0 mol
moles of CO2 = moles of butane × (coefficient of CO2 ÷ coefficient of butane)
moles of CO2 = 3.0 × (8 ÷ 2)
moles of CO2 = 12.0 mol
Part b. Calculate the moles of C₄H₁₀ that must burn completely to make 20.0 mol of H₂O. Show the same working.
Model: moles of water = 20.0 mol (the target)
moles of butane = moles of water × (coefficient of butane ÷ coefficient of water)
moles of butane = 20.0 × (2 ÷ 10)
moles of butane = 4.0 mol
Part c. For part (b), apply the lesson's sanity check without recalculating: compare the coefficient of the wanted substance with the coefficient of the given substance, state whether the answer should come out smaller or larger than the given moles, and confirm that your part (b) answer passes the check.
Model: The taught check says: when the wanted coefficient is smaller than the given one, the answer comes out smaller than the given moles. In part (b) the wanted substance butane has coefficient 2 and the given substance water has coefficient 10 — the wanted coefficient is smaller, so the answer should come out smaller than 20.0 mol. The answer 4.0 mol is smaller than 20.0 mol, so it passes the check.
Full model answer. (a)
moles of butane = 3.0 mol
moles of CO2 = moles of butane × (coefficient of CO2 ÷ coefficient of butane)
moles of CO2 = 3.0 × (8 ÷ 2)
moles of CO2 = 12.0 mol
(b)
moles of water = 20.0 mol (the target)
moles of butane = moles of water × (coefficient of butane ÷ coefficient of water)
moles of butane = 20.0 × (2 ÷ 10)
moles of butane = 4.0 mol
(c)
The taught check says: when the wanted coefficient is smaller than the given one, the answer comes out smaller than the given moles. In part (b) the wanted substance butane has coefficient 2 and the given substance water has coefficient 10 — the wanted coefficient is smaller, so the answer should come out smaller than 20.0 mol. The answer 4.0 mol is smaller than 20.0 mol, so it passes the check.
STO-009 · hard · application · seq 22 · authored (FRQ2 per composition.frq_split_rationale: STO-009 capstone chain end-to-end — plan + full mass-to-mass run + step-3 molar-mass rule; forward reactant→product only per the STO-010 boundary; both Ms supplied, corpus-free; fresh reaction + masses per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
K reacts with S to form K₂S as the balanced equation shows 2K + S → K₂S. 234.6 g of K reacts completely. (M of K = 39.1 g/mol; M of K₂S = 110.3 g/mol.)
K reacts with S to form K₂S as the balanced equation shows 2K + S → K₂S. 234.6 g of K reacts completely. (M of K = 39.1 g/mol; M of K₂S = 110.3 g/mol.)
(a) Before calculating, state the three-step plan for working out the mass of K₂S that forms, naming the tool each step uses.
(b) Carry out the calculation, showing the working for every step, and give the mass of K₂S that forms.
(c) State which substance's molar mass belongs in step 3, and why.
Part a. Before calculating, state the three-step plan for working out the mass of K₂S that forms, naming the tool each step uses.
Model: grams of K → moles of K (n = m/M) → moles of the potassium sulfide (mole-ratio equation) → grams of the potassium sulfide (m = n × M).
Part b. Carry out the calculation, showing the working for every step, and give the mass of K₂S that forms.
Model: m of K = 234.6 g; M of K = 39.1 g/mol; M of the potassium sulfide = 110.3 g/mol
Step 1 — convert the mass to moles:
n = m / M
n of K = 234.6 / 39.1
n of K = 6.0 mol
Step 2 — convert to moles of the potassium sulfide with the mole ratio:
moles of the potassium sulfide = moles of K × (coefficient of the potassium sulfide ÷ coefficient of K)
moles of the potassium sulfide = 6.0 × (1 ÷ 2)
moles of the potassium sulfide = 3.0 mol
Step 3 — convert the moles to mass:
m = n × M
m of the potassium sulfide = 3.0 × 110.3
m of the potassium sulfide = 330.9 g
Part c. State which substance's molar mass belongs in step 3, and why.
Model: Step 3 must use the WANTED substance's molar mass — the potassium sulfide's, 110.3 g/mol — because step 3 converts the wanted substance's moles into its mass with m = n × M. The given substance's molar mass (K's 39.1 g/mol) is used only in step 1 and never appears in step 3.
Full model answer. (a)
grams of K → moles of K (n = m/M) → moles of the potassium sulfide (mole-ratio equation) → grams of the potassium sulfide (m = n × M).
(b)
m of K = 234.6 g; M of K = 39.1 g/mol; M of the potassium sulfide = 110.3 g/mol
Step 1 — convert the mass to moles:
n = m / M
n of K = 234.6 / 39.1
n of K = 6.0 mol
Step 2 — convert to moles of the potassium sulfide with the mole ratio:
moles of the potassium sulfide = moles of K × (coefficient of the potassium sulfide ÷ coefficient of K)
moles of the potassium sulfide = 6.0 × (1 ÷ 2)
moles of the potassium sulfide = 3.0 mol
Step 3 — convert the moles to mass:
m = n × M
m of the potassium sulfide = 3.0 × 110.3
m of the potassium sulfide = 330.9 g
(c)
Step 3 must use the WANTED substance's molar mass — the potassium sulfide's, 110.3 g/mol — because step 3 converts the wanted substance's moles into its mass with m = n × M. The given substance's molar mass (K's 39.1 g/mol) is used only in step 1 and never appears in step 3.
STO-013 · hard · transfer · seq 23 · authored (FRQ3 per composition.frq_split_rationale: STO-013 inspection PRODUCTION — one planted error; D-form class = step-2 wrong-partner coefficient (O2's coefficient grabbed instead of KNO2's); fresh reaction + numbers per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
KNO₃ breaks down on heating: 2KNO₃ → 2KNO₂ + O₂. A student calculates the mass of KNO₂ formed when 202.2 g of KNO₃ breaks down completely. (M of KNO₃ = 101.1 g/mol; M of KNO₂ = 85.1 g/mol.)
Step 1: n of KNO₃ = 202.2 / 101.1 = 2.0 mol
Step 2: moles of KNO₂ = 2.0 × (1 ÷ 2) = 1.0 mol
Step 3: m of KNO₂ = 1.0 × 85.1 = 85.1 g
Exactly one step contains an error.
KNO₃ breaks down on heating: 2KNO₃ → 2KNO₂ + O₂. A student calculates the mass of KNO₂ formed when 202.2 g of KNO₃ breaks down completely. (M of KNO₃ = 101.1 g/mol; M of KNO₂ = 85.1 g/mol.)
Step 1: n of KNO₃ = 202.2 / 101.1 = 2.0 mol
Step 2: moles of KNO₂ = 2.0 × (1 ÷ 2) = 1.0 mol
Step 3: m of KNO₂ = 1.0 × 85.1 = 85.1 g
Exactly one step contains an error.
(a) Name the step that contains the error, and state what is wrong with it.
(b) Correct the erroneous step, rerun the working from that point, and give the corrected mass of KNO₂.
(c) A classmate claims step 3 must also contain an error, because it produced a wrong mass. Using the lesson's rule about steps that inherit a wrong number, state whether the classmate is right, and explain.
Part a. Name the step that contains the error, and state what is wrong with it.
Model: The error is in step 2. Checking each step: step 1 divides 202.2 g by KNO3's own molar mass, 101.1 g/mol — correct. In step 2 the mole ratio must be coefficient of KNO2 ÷ coefficient of KNO3 = 2 ÷ 2, read from the balanced equation, but the student put the coefficient of oxygen — the wrong partner — on top and wrote 1 ÷ 2.
Part b. Correct the erroneous step, rerun the working from that point, and give the corrected mass of KNO₂.
Model: Corrected step 2:
moles of KNO3 = 2.0 mol
moles of KNO2 = moles of KNO3 × (coefficient of KNO2 ÷ coefficient of KNO3)
moles of KNO2 = 2.0 × (2 ÷ 2)
moles of KNO2 = 2.0 mol
Rerun step 3:
m = n × M
m of KNO2 = 2.0 × 85.1
m of KNO2 = 170.2 g
Part c. A classmate claims step 3 must also contain an error, because it produced a wrong mass. Using the lesson's rule about steps that inherit a wrong number, state whether the classmate is right, and explain.
Model: Step 3 does not contain an error. The taught rule: a step that inherits a wrong number from an earlier step is not itself the error — the error lives where the wrong number was made. Step 3 used the correct method and the correct molar mass (KNO2's 85.1 g/mol); it only inherited the wrong 1.0 mol that step 2 made, so the error lives in step 2, not step 3.
Full model answer. (a)
The error is in step 2. Checking each step: step 1 divides 202.2 g by KNO3's own molar mass, 101.1 g/mol — correct. In step 2 the mole ratio must be coefficient of KNO2 ÷ coefficient of KNO3 = 2 ÷ 2, read from the balanced equation, but the student put the coefficient of oxygen — the wrong partner — on top and wrote 1 ÷ 2.
(b)
Corrected step 2:
moles of KNO3 = 2.0 mol
moles of KNO2 = moles of KNO3 × (coefficient of KNO2 ÷ coefficient of KNO3)
moles of KNO2 = 2.0 × (2 ÷ 2)
moles of KNO2 = 2.0 mol
Rerun step 3:
m = n × M
m of KNO2 = 2.0 × 85.1
m of KNO2 = 170.2 g
(c)
Step 3 does not contain an error. The taught rule: a step that inherits a wrong number from an earlier step is not itself the error — the error lives where the wrong number was made. Step 3 used the correct method and the correct molar mass (KNO2's 85.1 g/mol); it only inherited the wrong 1.0 mol that step 2 made, so the error lives in step 2, not step 3.
In Ti + 2I₂ → TiI₄, 2 molecules of I₂ react for every 1 atom of Ti. Why does the same two-to-one ratio hold for moles of I₂ and moles of Ti?
Because a mole is the same fixed count for every substance, so scaling every substance up by that same count leaves the ratio unchanged. KEY
Because 2 g of I₂ react for every 1 g of Ti, and whatever holds for the gram ratio carries straight over to the mole ratio.
Because I₂ and Ti have the same molar mass, so equal counts of them always weigh the same.
Because the mole is chosen fresh for each reaction so that the coefficients always describe it correctly.
STO-001 · easy · recall · seq 2 · authored (fresh chlorination with both coefficients written, coprime as asked; wave-5 fresh instance 2Fe + 3Cl₂ → 2FeCl₃; asked order Cl₂-first preserves the 3:2 answer contract) · text
In the balanced equation 2Fe + 3Cl₂ → 2FeCl₃, what is the mole ratio of Cl₂ to Fe? Write the ratio with a colon, such as 4:1.
Answer: 3:2 KEY
also accepted: 3 : 2, 3 to 2
wrongValueDiagnostics
[
{
"value": "2:3",
"misconception": "you reversed the order — the question asks Cl₂ to Fe, so Cl₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order. The question asks Cl₂ to Fe, so Cl₂'s coefficient 3 comes first: 3:2."
},
{
"value": "3:1",
"misconception": "you read the subscripts inside FeCl₃ — the 3 Cl and 1 Fe count atoms in one formula unit, not moles in the ratio",
"feedback": "This answer is wrong — you read the subscripts inside FeCl₃, which count atoms in one formula unit. The mole ratio comes from the written coefficients: 3 for Cl₂ and 2 for Fe, so the ratio is 3:2."
},
{
"value": "1:1",
"misconception": "you ignored the coefficients — the mole ratio is read from the written coefficients, 3 for Cl₂ and 2 for Fe",
"feedback": "This answer is wrong — you ignored the coefficients. The mole ratio comes from the written coefficients: 3 for Cl₂ and 2 for Fe, so the ratio is 3:2."
}
]
feedback
{
"keyExplanation": "The coefficient of Cl₂ is 3, and the coefficient of Fe is 2. Take the coefficients in the order asked: Cl₂ first, Fe second. The subscripts inside FeCl₃ play no part. The mole ratio of Cl₂ to Fe is 3:2.",
"anticipatedWrong": [
{
"value": "2:3",
"misconception": "you reversed the order — the question asks Cl₂ to Fe, so Cl₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order. The question asks Cl₂ to Fe, so Cl₂'s coefficient 3 comes first: 3:2."
},
{
"value": "3:1",
"misconception": "you read the subscripts inside FeCl₃ — the 3 Cl and 1 Fe count atoms in one formula unit, not moles in the ratio",
"feedback": "This answer is wrong — you read the subscripts inside FeCl₃, which count atoms in one formula unit. The mole ratio comes from the written coefficients: 3 for Cl₂ and 2 for Fe, so the ratio is 3:2."
},
{
"value": "1:1",
"misconception": "you ignored the coefficients — the mole ratio is read from the written coefficients, 3 for Cl₂ and 2 for Fe",
"feedback": "This answer is wrong — you ignored the coefficients. The mole ratio comes from the written coefficients: 3 for Cl₂ and 2 for Fe, so the ratio is 3:2."
}
]
}
STO-003 · easy · recall · seq 3 · authored (same miscounted-atoms reason read against a fresh unbalanced-vs-balanced pair, kept qualitative; fresh instance Ti + N₂ → TiN vs 2Ti + N₂ → 2TiN; foils = small-amounts-ok / direction / fixed-factor) · mcq4
Two versions of an equation are on the board: the unbalanced Ti + N₂ → TiN and the balanced 2Ti + N₂ → 2TiN. Why can a mole ratio be read only from the balanced version?
Only the balanced coefficients count atoms correctly — the unbalanced version shows atoms disappearing, and a reaction only rearranges atoms. KEY
The unbalanced version also gives a perfectly usable ratio, provided that only a small amount of the Ti is actually reacting.
The unbalanced version lists the product among the reactants, so its ratio pairs the wrong substances.
The unbalanced version understates every coefficient by the same fixed factor, so scaling it up would also work.
STO-006 · easy · recall · seq 4 · authored (step-order rotation per blueprint — full-sequence read, instance-free, zero arithmetic per JM-P1a) · mcq4
Which sequence puts the three steps of a mass-to-mass calculation in the correct order?
Mass to moles, then the mole ratio, then moles to mass. KEY
The mole ratio, then mass to moles, then moles to mass.
Mass to moles, then moles to mass, then the mole ratio.
Moles to mass, then the mole ratio, then mass to moles.
STO-004 · medium · application · seq 5 · authored (parallel mcq slot; fresh reverse setup on a nonmetal-oxide synthesis, all three coefficients distinct) · mcq4
For 4P + 5O₂ → 2P₂O₅, which equation finds the moles of O₂ needed from the moles of P₂O₅?
moles of O₂ = moles of P₂O₅ × (5 ÷ 2) KEY
moles of O₂ = moles of P₂O₅ × (2 ÷ 5)
moles of O₂ = moles of P₂O₅ × (4 ÷ 2)
moles of O₂ = moles of P₂O₅ × (5 ÷ 4)
STO-005 · medium · application · seq 6 · authored (parallel numeric slot; fresh burn surface given from the O₂ side, integer given moles, exact 7:11 ratio) · numeric
In C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O, 11.0 mol of O₂ reacts completely with C₇H₁₆. How many moles of CO₂ form? Give your answer in moles to one decimal place.
Answer: 7.0 mol KEY
also accepted: 7.0 mol
wrongValueDiagnostics
[
{
"value": 17.3,
"misconception": "you turned the fraction upside down — the wanted substance CO₂'s coefficient 7 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance CO₂'s coefficient 7 sits on top."
},
{
"value": 11.0,
"misconception": "you ignored the coefficients and matched the moles one-for-one — the ratio of CO₂ to O₂ is 7:11",
"feedback": "This answer is wrong — you ignored the coefficients and matched the moles one-for-one — the ratio of CO₂ to O₂ is 7:11."
},
{
"value": 77.0,
"misconception": "you multiplied by CO₂'s coefficient alone instead of the coefficient ratio",
"feedback": "This answer is wrong — you multiplied by CO₂'s coefficient alone instead of the coefficient ratio."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of O₂ = 11.0 mol\nWrite down the equation:\nmoles of CO₂ = moles of O₂ × (coefficient of CO₂ ÷ coefficient of O₂)\nSubstitute in the values, and calculate:\nmoles of CO₂ = 11.0 × (7 ÷ 11)\nmoles of CO₂ = 7.0 mol"
}
STO-001 · medium · application · seq 7 · authored (parallel text slot; fresh unwritten-1 pair with the inside-subscript trap on the asked substance Ca₃P₂; wave-6 fresh instance 3Ca + 2P → Ca₃P₂) · text
In the balanced equation 3Ca + 2P → Ca₃P₂, what is the mole ratio of Ca₃P₂ to Ca? Write the ratio with a colon, such as 4:1.
Answer: 1:3 KEY
also accepted: 1 : 3, 1 to 3
wrongValueDiagnostics
[
{
"value": "3:3",
"misconception": "you used the subscript 3 inside Ca₃P₂ as its coefficient — Ca₃P₂ has no written coefficient, so its coefficient is 1",
"feedback": "This answer is wrong — you used the subscript 3 inside Ca₃P₂ as its coefficient — Ca₃P₂ has no written coefficient, so its coefficient is 1."
},
{
"value": "3:1",
"misconception": "you reversed the order — the question asks Ca₃P₂ to Ca, so Ca₃P₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order — the question asks Ca₃P₂ to Ca, so Ca₃P₂'s coefficient comes first."
},
{
"value": "1:2",
"misconception": "you took P's coefficient for Ca — the second number must belong to Ca, whose coefficient is 3",
"feedback": "This answer is wrong — you took P's coefficient for Ca — the second number must belong to Ca, whose coefficient is 3."
}
]
feedback
{
"keyExplanation": "Ca₃P₂ has no written coefficient, so its coefficient is 1; the coefficient of Ca is 3. The subscript 3 inside Ca₃P₂ counts atoms in one formula unit and plays no part in the mole ratio. The mole ratio of Ca₃P₂ to Ca is 1:3.",
"anticipatedWrong": [
{
"value": "3:3",
"misconception": "you used the subscript 3 inside Ca₃P₂ as its coefficient — Ca₃P₂ has no written coefficient, so its coefficient is 1",
"feedback": "This answer is wrong — you used the subscript 3 inside Ca₃P₂ as its coefficient — Ca₃P₂ has no written coefficient, so its coefficient is 1."
},
{
"value": "3:1",
"misconception": "you reversed the order — the question asks Ca₃P₂ to Ca, so Ca₃P₂'s coefficient comes first",
"feedback": "This answer is wrong — you reversed the order — the question asks Ca₃P₂ to Ca, so Ca₃P₂'s coefficient comes first."
},
{
"value": "1:2",
"misconception": "you took P's coefficient for Ca — the second number must belong to Ca, whose coefficient is 3",
"feedback": "This answer is wrong — you took P's coefficient for Ca — the second number must belong to Ca, whose coefficient is 3."
}
]
}
STO-007 · medium · application · seq 8 · authored (parallel numeric slot; fresh decomposition surface per the case rotation; only the given substance's M supplied) · numeric
In 2HgO → 2Hg + O₂, 324.9 g of HgO (M = 216.6 g/mol) breaks down completely. How many moles of O₂ are released? Give your answer in moles to 2 significant figures.
Answer: 0.75 mol KEY
also accepted: 0.75 mol
wrongValueDiagnostics
[
{
"value": 3.0,
"misconception": "you turned the fraction upside down — the wanted substance O₂'s coefficient 1 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance O₂'s coefficient 1 sits on top."
},
{
"value": 1.5,
"misconception": "you skipped the mole-ratio step — 1.5 mol is the HgO's moles, not the O₂'s",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 1.5 mol is the HgO's moles, not the O₂'s."
},
{
"value": 162.45,
"misconception": "you applied the mole ratio to the mass instead of converting to moles first",
"feedback": "This answer is wrong — you applied the mole ratio to the mass instead of converting to moles first."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of HgO = 324.9 g, M of HgO = 216.6 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of HgO = 324.9 / 216.6\nn of HgO = 1.5 mol\nStep 2 — convert to moles of O₂ with the mole ratio:\nmoles of O₂ = moles of HgO × (coefficient of O₂ ÷ coefficient of HgO)\nmoles of O₂ = 1.5 × (1 ÷ 2)\nmoles of O₂ = 0.75 mol"
}
STO-008 · medium · application · seq 9 · authored (parallel numeric slot; fresh 2:3 awkward-ratio instance, P per the allocation's allowed fresh-element set) · numeric
In 2P + 3Br₂ → 2PBr₃, 6.0 mol of Br₂ reacts completely with P. What mass of PBr₃ forms? (M of PBr₃ = 270.7 g/mol) Give your answer in grams to one decimal place.
Answer: 1082.8 g KEY
also accepted: 1082.8 g, 1,082.8
wrongValueDiagnostics
[
{
"value": 1624.2,
"misconception": "you skipped the mole-ratio step — 6.0 mol is the Br₂'s moles, not the PBr₃'s",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 6.0 mol is the Br₂'s moles, not the PBr₃'s."
},
{
"value": 2436.3,
"misconception": "you turned the fraction upside down — the wanted substance PBr₃'s coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance PBr₃'s coefficient 2 sits on top."
},
{
"value": 639.2,
"misconception": "you used Br₂'s molar mass (159.8 g/mol) — step 2 needs the WANTED substance PBr₃'s molar mass",
"feedback": "This answer is wrong — you used Br₂'s molar mass (159.8 g/mol) — step 2 needs the WANTED substance PBr₃'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of Br₂ = 6.0 mol, M of PBr₃ = 270.7 g/mol\nStep 1 — convert to moles of PBr₃ with the mole ratio:\nmoles of PBr₃ = moles of Br₂ × (coefficient of PBr₃ ÷ coefficient of Br₂)\nmoles of PBr₃ = 6.0 × (2 ÷ 3)\nmoles of PBr₃ = 4.0 mol\nStep 2 — convert the moles to mass:\nm = n × M\nm of PBr₃ = 4.0 × 270.7\nm of PBr₃ = 1082.8 g"
}
STO-012 · medium · application · seq 10 · authored (prediction-class rotation on a fresh fuel — hexane; completes the blueprint's defense/prediction/collected-products rotation: A defense, B prediction-metal, C collected-products, D defense-metal, E prediction-fuel) · mcq4
A lab burner uses 43.0 g of hexane, C₆H₁₄, as its fuel. Predict how the combined mass of ALL the CO₂ and H₂O the flame produces will compare with the 43.0 g of hexane, and why.
Greater — the O₂ that reacts adds its own mass to the products. KEY
The same — mass is conserved, so the products must weigh exactly what the hexane weighed.
Less — burning destroys some of the fuel's mass as heat and light.
Greater — hot gases weigh more than the cool liquid fuel they came from.
STO-005 · medium · application · seq 11 · authored (fresh O2-needed metal-oxide instance, 3:2 ratio; 2W+3O2->2WO3 equation is bank-fresh — existing rows use only W+3F2) · numeric
In 2W + 3O₂ → 2WO₃, how many moles of O₂ are needed to react completely with 2.0 mol of W? Give your answer in moles to one decimal place.
Answer: 3.0 mol KEY
also accepted: 3.0 mol
wrongValueDiagnostics
[
{
"value": 1.3333,
"misconception": "you turned the fraction upside down — the wanted substance O₂'s coefficient 3 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance O₂'s coefficient 3 sits on top."
},
{
"value": 2.0,
"misconception": "you ignored the coefficients and matched the moles one-for-one — the ratio of O₂ to W is 3:2",
"feedback": "This answer is wrong — you ignored the coefficients and matched the moles one-for-one — the ratio of O₂ to W is 3:2."
},
{
"value": 6.0,
"misconception": "you multiplied by O₂'s coefficient alone instead of the coefficient ratio",
"feedback": "This answer is wrong — you multiplied by O₂'s coefficient alone instead of the coefficient ratio."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of W = 2.0 mol\nWrite down the equation:\nmoles of O₂ = moles of W × (coefficient of O₂ ÷ coefficient of W)\nSubstitute in the values, and calculate:\nmoles of O₂ = 2.0 × (3 ÷ 2)\nmoles of O₂ = 3.0 mol"
}
STO-009 · medium · application · seq 12 · authored (parallel numeric slot; fresh 2:4 metal-oxide chain — Sb2O3 equation bank-fresh, existing rows use only 2Sb+3Cl2) · numeric
In 4Sb + 3O₂ → 2Sb₂O₃, 730.8 g of Sb (M = 121.8 g/mol) burns completely. What mass of Sb₂O₃ forms? (M of Sb₂O₃ = 291.6 g/mol) Give your answer in grams to one decimal place.
Answer: 874.8 g KEY
also accepted: 874.8 g
wrongValueDiagnostics
[
{
"value": 1749.6,
"misconception": "you skipped the mole-ratio step — 6.0 mol is the Sb's moles, and only half as many moles of Sb₂O₃ form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 6.0 mol is the Sb's moles, and only half as many moles of Sb₂O₃ form."
},
{
"value": 3499.2,
"misconception": "you turned the fraction upside down — the wanted substance Sb₂O₃'s coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Sb₂O₃'s coefficient 2 sits on top."
},
{
"value": 365.4,
"misconception": "you used Sb's molar mass in step 3 — the final step needs the WANTED substance Sb₂O₃'s molar mass",
"feedback": "This answer is wrong — you used Sb's molar mass in step 3 — the final step needs the WANTED substance Sb₂O₃'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Sb = 730.8 g, M of Sb = 121.8 g/mol, M of Sb₂O₃ = 291.6 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Sb = 730.8 / 121.8\nn of Sb = 6.0 mol\nStep 2 — convert to moles of Sb₂O₃ with the mole ratio:\nmoles of Sb₂O₃ = moles of Sb × (coefficient of Sb₂O₃ ÷ coefficient of Sb)\nmoles of Sb₂O₃ = 6.0 × (2 ÷ 4)\nmoles of Sb₂O₃ = 3.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of Sb₂O₃ = 3.0 × 291.6\nm of Sb₂O₃ = 874.8 g"
}
Ag reacts completely with Cl₂: 2Ag + Cl₂ → 2AgCl. (M of Ag = 107.9 g/mol, M of Cl₂ = 71.0 g/mol.) What total mass of reactants does the balanced equation represent? Give your answer in grams to one decimal place.
Answer: 286.8 g KEY
also accepted: 286.8 g
wrongValueDiagnostics
[
{
"value": 178.9,
"misconception": "you ignored both coefficients — each substance's mass is its coefficient multiplied by its molar mass",
"feedback": "This answer is wrong — you ignored both coefficients — each substance's mass is its coefficient multiplied by its molar mass."
},
{
"value": 249.9,
"misconception": "you swapped the coefficients — the 2 belongs to Ag and the unwritten 1 belongs to Cl₂",
"feedback": "This answer is wrong — you swapped the coefficients — the 2 belongs to Ag and the unwritten 1 belongs to Cl₂."
},
{
"value": 181.9,
"misconception": "you added each coefficient instead of multiplying by it",
"feedback": "This answer is wrong — you added each coefficient instead of multiplying by it."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM of Ag = 107.9 g/mol, M of Cl₂ = 71.0 g/mol\nWrite down the equation:\nmass of each substance = coefficient × molar mass\nSubstitute in the values, and calculate:\ntotal mass of reactants = 2 × 107.9 + 1 × 71.0\ntotal mass of reactants = 286.8 g"
}
STO-007 · medium · application · seq 14 · authored (parallel numeric slot; reactant-mass-given read on a fresh glucose-respiration surface C6H12O6 + 6O2 -> 6CO2 + 6H2O; unwritten coefficient 1 on the given substance) · numeric
In a respiration demonstration, glucose reacts completely: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. 99.0 g of glucose, C₆H₁₂O₆ (M = 180.0 g/mol), is used up. How many moles of CO₂ form? Give your answer in moles to one decimal place.
Answer: 3.3 mol KEY
also accepted: 3.3 mol
wrongValueDiagnostics
[
{
"value": 0.092,
"misconception": "you turned the fraction upside down — the wanted substance CO₂'s coefficient 6 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance CO₂'s coefficient 6 sits on top."
},
{
"value": 0.55,
"misconception": "you stopped at step 1 — 0.55 mol is the glucose's moles, not the CO₂'s",
"feedback": "This answer is wrong — you stopped at step 1 — 0.55 mol is the glucose's moles, not the CO₂'s."
},
{
"value": 594.0,
"misconception": "you applied the mole ratio to the mass instead of converting to moles first",
"feedback": "This answer is wrong — you applied the mole ratio to the mass instead of converting to moles first."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of C₆H₁₂O₆ = 99.0 g, M of C₆H₁₂O₆ = 180.0 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of C₆H₁₂O₆ = 99.0 / 180.0\nn of C₆H₁₂O₆ = 0.55 mol\nStep 2 — convert to moles of CO₂ with the mole ratio:\nmoles of CO₂ = moles of C₆H₁₂O₆ × (coefficient of CO₂ ÷ coefficient of C₆H₁₂O₆)\nmoles of CO₂ = 0.55 × (6 ÷ 1)\nmoles of CO₂ = 3.3 mol"
}
In 4V + 3O₂ → 2V₂O₃, what mass of V must react completely with O₂ to make 524.3 g of V₂O₃? (M of V₂O₃ = 149.8 g/mol; M of V = 50.9 g/mol) Give your answer in grams to one decimal place.
Answer: 356.3 g KEY
also accepted: 356.3 g
wrongValueDiagnostics
[
{
"value": 178.2,
"misconception": "you skipped the mole-ratio step — 3.5 mol is the V₂O₃'s moles, and twice as many moles of V are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 3.5 mol is the V₂O₃'s moles, and twice as many moles of V are needed."
},
{
"value": 89.1,
"misconception": "you turned the fraction upside down — the wanted substance V's coefficient 4 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance V's coefficient 4 sits on top."
},
{
"value": 1048.6,
"misconception": "you used V₂O₃'s molar mass in step 3 — the final step needs the WANTED substance V's molar mass",
"feedback": "This answer is wrong — you used V₂O₃'s molar mass in step 3 — the final step needs the WANTED substance V's molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of V₂O₃ = 524.3 g, M of V₂O₃ = 149.8 g/mol, M of V = 50.9 g/mol\nStep 1 — convert the target mass to moles:\nn = m / M\nn of V₂O₃ = 524.3 / 149.8\nn of V₂O₃ = 3.5 mol\nStep 2 — convert to moles of V with the mole ratio:\nmoles of V = moles of V₂O₃ × (coefficient of V ÷ coefficient of V₂O₃)\nmoles of V = 3.5 × (4 ÷ 2)\nmoles of V = 7.0 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of V = 7.0 × 50.9\nm of V = 356.3 g"
}
STO-008 · medium · application · seq 16 · fully authored (F3 reverse pairing on a fresh lab-decomposition surface 2PbO2 -> 2PbO + O2; reactant mass needed to release an integer target moles of O2; the withheld corpus-banned M(O2)=32.0 appears ONLY inside the wrong-M misconception feedback per specNotes convention) · numeric
In 2PbO₂ → 2PbO + O₂, what mass of PbO₂ must break down completely to release 5.0 mol of O₂? (M of PbO₂ = 239.2 g/mol) Give your answer in grams to one decimal place.
Answer: 2392.0 g KEY
also accepted: 2392.0 g
wrongValueDiagnostics
[
{
"value": 1196.0,
"misconception": "you skipped the mole-ratio step — 5.0 mol is the O₂'s moles, and twice as many moles of PbO₂ are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 5.0 mol is the O₂'s moles, and twice as many moles of PbO₂ are needed."
},
{
"value": 598.0,
"misconception": "you turned the fraction upside down — the wanted substance PbO₂'s coefficient 2 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance PbO₂'s coefficient 2 sits on top."
},
{
"value": 320.0,
"misconception": "you used O₂'s molar mass (32.0 g/mol) — step 2 needs the WANTED substance PbO₂'s molar mass",
"feedback": "This answer is wrong — you used O₂'s molar mass (32.0 g/mol) — step 2 needs the WANTED substance PbO₂'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of O₂ = 5.0 mol, M of PbO₂ = 239.2 g/mol\nStep 1 — convert to moles of PbO₂ with the mole ratio:\nmoles of PbO₂ = moles of O₂ × (coefficient of PbO₂ ÷ coefficient of O₂)\nmoles of PbO₂ = 5.0 × (2 ÷ 1)\nmoles of PbO₂ = 10.0 mol\nStep 2 — convert the moles to mass:\nm = n × M\nm of PbO₂ = 10.0 × 239.2\nm of PbO₂ = 2392.0 g"
}
STO-011 · hard · transfer · seq 17 · authored (fresh audit per blueprint case rotation; error class: subscript read as a coefficient) · mcq4
For Ba + I₂ → BaI₂, a student totals the two sides and gets 391.1 g of reactants but 782.2 g of products. (M of Ba = 137.3 g/mol, M of I₂ = 253.8 g/mol, M of BaI₂ = 391.1 g/mol.) What went wrong?
The subscript 2 in BaI₂ was read as a coefficient — BaI₂'s coefficient is 1, so the product side is 1 × 391.1 = 391.1 g. KEY
The reactant total is wrong — the I₂ needs a coefficient of 2, because each BaI₂ holds two iodine atoms that must both be supplied.
The equation itself is unbalanced, so the totals cannot match.
Nothing went wrong — the product always outweighs the reactants that form it.
STO-013 · hard · transfer · seq 18 · authored (fresh instance; error class per blueprint rotation: step-2 wrong-partner-coefficient) · mcq4
For 2Sr + O₂ → 2SrO, a student calculates the mass of SrO formed from 306.6 g of Sr. (M of Sr = 87.6 g/mol, M of SrO = 103.6 g/mol.)
Step 1: n of Sr = 306.6 / 87.6 = 3.5 mol
Step 2: moles of SrO = 3.5 × 1/2 = 1.75 mol
Step 3: m of SrO = 1.75 × 103.6 = 181.3 g
Which step contains the error?
Step 2 KEY
Step 1
Step 3
No step contains an error
STO-010 · hard · transfer · seq 19 · fully authored per blueprint (production-planning framing, awkward 3:2 pair, both Ms supplied; wrongs: skipped ratio / inverted / wrong-M) · numeric
A catalyst supplier must produce 875.1 g of CrBr₃ in one batch. In 2Cr + 3Br₂ → 2CrBr₃, what mass of Br₂ must react completely to supply the batch? (M of CrBr₃ = 291.7 g/mol; M of Br₂ = 159.8 g/mol) Give your answer in grams to one decimal place.
Answer: 719.1 g KEY
also accepted: 719.1 g
wrongValueDiagnostics
[
{
"value": 479.4,
"misconception": "you skipped the mole-ratio step — 3.0 mol is the CrBr₃'s moles, and half as many again moles of Br₂ are needed",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 3.0 mol is the CrBr₃'s moles, and half as many again moles of Br₂ are needed."
},
{
"value": 319.6,
"misconception": "you turned the fraction upside down — the wanted substance Br₂'s coefficient 3 sits on top",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Br₂'s coefficient 3 sits on top."
},
{
"value": 1312.7,
"misconception": "you used CrBr₃'s molar mass in step 3 — the final step needs the WANTED substance Br₂'s molar mass",
"feedback": "This answer is wrong — you used CrBr₃'s molar mass in step 3 — the final step needs the WANTED substance Br₂'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of CrBr₃ = 875.1 g, M of CrBr₃ = 291.7 g/mol, M of Br₂ = 159.8 g/mol\nStep 1 — convert the target mass to moles:\nn = m / M\nn of CrBr₃ = 875.1 / 291.7\nn of CrBr₃ = 3.0 mol\nStep 2 — convert to moles of Br₂ with the mole ratio:\nmoles of Br₂ = moles of CrBr₃ × (coefficient of Br₂ ÷ coefficient of CrBr₃)\nmoles of Br₂ = 3.0 × (3 ÷ 2)\nmoles of Br₂ = 4.5 mol\nStep 3 — convert the moles to mass:\nm = n × M\nm of Br₂ = 4.5 × 159.8\nm of Br₂ = 719.1 g"
}
STO-009 · hard · transfer · seq 20 · authored (parallel forward mass-to-mass slot) | stage-3 re-instance: the original decomposition chain duplicated A#8's reaction -> 3Mg+N₂ nitride chain (census-fresh) · numeric
A metals plant burns magnesium in a flow of N₂: in 3Mg + N₂ → Mg₃N₂, 145.8 g of Mg (M of Mg = 24.3 g/mol) reacts completely. What mass of Mg₃N₂ forms? (M of Mg₃N₂ = 100.9 g/mol) Give your answer in grams to one decimal place.
Answer: 201.8 g KEY
also accepted: 201.8 g
wrongValueDiagnostics
[
{
"value": 605.4,
"misconception": "you skipped the mole-ratio step — 6.0 mol is the Mg's moles; only a third as many moles of Mg₃N₂ form",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 6.0 mol is the Mg's moles; only a third as many moles of Mg₃N₂ form."
},
{
"value": 1816.2,
"misconception": "you turned the fraction upside down — the wanted substance Mg₃N₂'s coefficient 1 sits on top, over Mg's 3",
"feedback": "This answer is wrong — you turned the fraction upside down — the wanted substance Mg₃N₂'s coefficient 1 sits on top, over Mg's 3."
},
{
"value": 48.6,
"misconception": "you used Mg's molar mass in step 3 — the final step needs the WANTED substance Mg₃N₂'s molar mass",
"feedback": "This answer is wrong — you used Mg's molar mass in step 3 — the final step needs the WANTED substance Mg₃N₂'s molar mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Mg = 145.8 g, M of Mg = 24.3 g/mol, M of Mg₃N₂ = 100.9 g/mol\nStep 1 — convert the mass to moles:\nn = m / M\nn of Mg = 145.8 / 24.3\nn of Mg = 6.0 mol\nStep 2 — convert to moles of Mg₃N₂ with the mole ratio:\nmoles of Mg₃N₂ = moles of Mg × (coefficient of Mg₃N₂ ÷ coefficient of Mg)\nmoles of Mg₃N₂ = 6.0 × (1 ÷ 3)\nmoles of Mg₃N₂ = 2.0 mol\nStep 3 — convert the moles to a mass with the WANTED substance's molar mass:\nm = n × M\nm of Mg₃N₂ = 2.0 × 100.9\nm of Mg₃N₂ = 201.8 g"
}
STO-005 · medium · application · seq 21 · authored (FRQ1 per composition.frq_split_rationale: STO-005 mole-to-mole in BOTH roles + taught sanity check; mole-only, no molar masses per the STO-007+ boundary; fresh reaction + numbers per form (ZnS roasting); equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
ZnS is roasted in O₂ according to the balanced equation 2ZnS + 3O₂ → 2ZnO + 2SO₂. Work in moles throughout — no molar masses are needed.
ZnS is roasted in O₂ according to the balanced equation 2ZnS + 3O₂ → 2ZnO + 2SO₂. Work in moles throughout — no molar masses are needed.
(a) 9.0 mol of O₂ reacts completely with ZnS. Calculate the moles of ZnO that form. Show your working: state the given moles, write the mole-ratio equation with both substances named, substitute, and calculate.
(b) Calculate the moles of O₂ needed to make 2.0 mol of SO₂. Show the same working.
(c) For part (a), apply the lesson's sanity check without recalculating: compare the coefficient of the wanted substance with the coefficient of the given substance, state whether the answer should come out smaller or larger than the given moles, and confirm that your part (a) answer passes the check.
Part a. 9.0 mol of O₂ reacts completely with ZnS. Calculate the moles of ZnO that form. Show your working: state the given moles, write the mole-ratio equation with both substances named, substitute, and calculate.
Model: moles of oxygen = 9.0 mol
moles of ZnO = moles of oxygen × (coefficient of ZnO ÷ coefficient of oxygen)
moles of ZnO = 9.0 × (2 ÷ 3)
moles of ZnO = 6.0 mol
Part b. Calculate the moles of O₂ needed to make 2.0 mol of SO₂. Show the same working.
Model: moles of SO2 = 2.0 mol (the target)
moles of oxygen = moles of SO2 × (coefficient of oxygen ÷ coefficient of SO2)
moles of oxygen = 2.0 × (3 ÷ 2)
moles of oxygen = 3.0 mol
Part c. For part (a), apply the lesson's sanity check without recalculating: compare the coefficient of the wanted substance with the coefficient of the given substance, state whether the answer should come out smaller or larger than the given moles, and confirm that your part (a) answer passes the check.
Model: The taught check says: when the wanted coefficient is smaller than the given one, the answer comes out smaller than the given moles. In part (a) the wanted substance ZnO has coefficient 2 and the given substance oxygen has coefficient 3 — the wanted coefficient is smaller, so the answer should come out smaller than 9.0 mol. The answer 6.0 mol is smaller than 9.0 mol, so it passes the check.
Full model answer. (a)
moles of oxygen = 9.0 mol
moles of ZnO = moles of oxygen × (coefficient of ZnO ÷ coefficient of oxygen)
moles of ZnO = 9.0 × (2 ÷ 3)
moles of ZnO = 6.0 mol
(b)
moles of SO2 = 2.0 mol (the target)
moles of oxygen = moles of SO2 × (coefficient of oxygen ÷ coefficient of SO2)
moles of oxygen = 2.0 × (3 ÷ 2)
moles of oxygen = 3.0 mol
(c)
The taught check says: when the wanted coefficient is smaller than the given one, the answer comes out smaller than the given moles. In part (a) the wanted substance ZnO has coefficient 2 and the given substance oxygen has coefficient 3 — the wanted coefficient is smaller, so the answer should come out smaller than 9.0 mol. The answer 6.0 mol is smaller than 9.0 mol, so it passes the check.
STO-009 · hard · application · seq 22 · authored (FRQ2 per composition.frq_split_rationale: STO-009 capstone chain end-to-end — plan + full mass-to-mass run + step-3 molar-mass rule; forward reactant→product only per the STO-010 boundary; both Ms supplied, corpus-free (Br2 from the blueprint's halide-partner palette); fresh reaction + masses per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
Al reacts with Br₂ to form AlBr₃ as the balanced equation shows 2Al + 3Br₂ → 2AlBr₃. 479.4 g of Br₂ reacts completely with Al. (M of Br₂ = 159.8 g/mol; M of AlBr₃ = 266.7 g/mol.)
Al reacts with Br₂ to form AlBr₃ as the balanced equation shows 2Al + 3Br₂ → 2AlBr₃. 479.4 g of Br₂ reacts completely with Al. (M of Br₂ = 159.8 g/mol; M of AlBr₃ = 266.7 g/mol.)
(a) Before calculating, state the three-step plan for working out the mass of AlBr₃ that forms, naming the tool each step uses.
(b) Carry out the calculation, showing the working for every step, and give the mass of AlBr₃ that forms.
(c) State which substance's molar mass belongs in step 3, and why.
Part a. Before calculating, state the three-step plan for working out the mass of AlBr₃ that forms, naming the tool each step uses.
Model: grams of bromine → moles of bromine (n = m/M) → moles of AlBr3 (mole-ratio equation) → grams of AlBr3 (m = n × M).
Part b. Carry out the calculation, showing the working for every step, and give the mass of AlBr₃ that forms.
Model: m of bromine = 479.4 g; M of bromine = 159.8 g/mol; M of AlBr3 = 266.7 g/mol
Step 1 — convert the mass to moles:
n = m / M
n of bromine = 479.4 / 159.8
n of bromine = 3.0 mol
Step 2 — convert to moles of AlBr3 with the mole ratio:
moles of AlBr3 = moles of bromine × (coefficient of AlBr3 ÷ coefficient of bromine)
moles of AlBr3 = 3.0 × (2 ÷ 3)
moles of AlBr3 = 2.0 mol
Step 3 — convert the moles to mass:
m = n × M
m of AlBr3 = 2.0 × 266.7
m of AlBr3 = 533.4 g
Part c. State which substance's molar mass belongs in step 3, and why.
Model: Step 3 must use the WANTED substance's molar mass — AlBr3's, 266.7 g/mol — because step 3 converts the wanted substance's moles into its mass with m = n × M. The given substance's molar mass (bromine's 159.8 g/mol) is used only in step 1 and never appears in step 3.
Full model answer. (a)
grams of bromine → moles of bromine (n = m/M) → moles of AlBr3 (mole-ratio equation) → grams of AlBr3 (m = n × M).
(b)
m of bromine = 479.4 g; M of bromine = 159.8 g/mol; M of AlBr3 = 266.7 g/mol
Step 1 — convert the mass to moles:
n = m / M
n of bromine = 479.4 / 159.8
n of bromine = 3.0 mol
Step 2 — convert to moles of AlBr3 with the mole ratio:
moles of AlBr3 = moles of bromine × (coefficient of AlBr3 ÷ coefficient of bromine)
moles of AlBr3 = 3.0 × (2 ÷ 3)
moles of AlBr3 = 2.0 mol
Step 3 — convert the moles to mass:
m = n × M
m of AlBr3 = 2.0 × 266.7
m of AlBr3 = 533.4 g
(c)
Step 3 must use the WANTED substance's molar mass — AlBr3's, 266.7 g/mol — because step 3 converts the wanted substance's moles into its mass with m = n × M. The given substance's molar mass (bromine's 159.8 g/mol) is used only in step 1 and never appears in step 3.
STO-013 · hard · transfer · seq 23 · authored (FRQ3 per composition.frq_split_rationale: STO-013 inspection PRODUCTION — one planted error; E-form class = step-3 given-M (Zn's M reused, so the wrong mass equals the starting mass — the taught quiet-error tell); fresh reaction + numbers per form; equation uniqueness vs rows 1-20 to be verified at stage 3) · frq
For Zn + S → ZnS, a student calculates the mass of ZnS formed when 130.8 g of Zn reacts completely with S. (M of Zn = 65.4 g/mol; M of ZnS = 97.5 g/mol.)
Step 1: n of Zn = 130.8 / 65.4 = 2.0 mol
Step 2: moles of ZnS = 2.0 × (1 ÷ 1) = 2.0 mol
Step 3: m of ZnS = 2.0 × 65.4 = 130.8 g
Exactly one step contains an error.
For Zn + S → ZnS, a student calculates the mass of ZnS formed when 130.8 g of Zn reacts completely with S. (M of Zn = 65.4 g/mol; M of ZnS = 97.5 g/mol.)
Step 1: n of Zn = 130.8 / 65.4 = 2.0 mol
Step 2: moles of ZnS = 2.0 × (1 ÷ 1) = 2.0 mol
Step 3: m of ZnS = 2.0 × 65.4 = 130.8 g
Exactly one step contains an error.
(a) Name the step that contains the error, and state what is wrong with it.
(b) Correct the erroneous step and give the corrected mass of ZnS, showing the working.
(c) The final mass came out exactly equal to the starting mass, and a classmate argues the whole calculation must therefore be wrong from step 1 onward. Using the lesson's rule that an error lives where the wrong number was made, state which step made the wrong number and whether any step in this calculation merely inherited one.
Part a. Name the step that contains the error, and state what is wrong with it.
Model: The error is in step 3. Checking each step: step 1 divides 130.8 g by Zn's own molar mass, 65.4 g/mol — correct. Step 2 reads the ratio coefficient of ZnS ÷ coefficient of Zn = 1 ÷ 1 from the balanced equation — correct. Step 3 must use the WANTED substance's molar mass, ZnS's 97.5 g/mol, but the student used Zn's 65.4 g/mol — the given substance's molar mass. That is why the answer came out equal to the starting mass.
Part b. Correct the erroneous step and give the corrected mass of ZnS, showing the working.
Model: Corrected step 3:
moles of ZnS = 2.0 mol
m = n × M
m of ZnS = 2.0 × 97.5
m of ZnS = 195.0 g
Part c. The final mass came out exactly equal to the starting mass, and a classmate argues the whole calculation must therefore be wrong from step 1 onward. Using the lesson's rule that an error lives where the wrong number was made, state which step made the wrong number and whether any step in this calculation merely inherited one.
Model: The wrong number was made in step 3 itself: steps 1 and 2 both made correct numbers, so the classmate's claim that the calculation is wrong from step 1 onward fails. No step in this calculation merely inherited a wrong number — nothing comes after step 3. The error lives in step 3, where the wrong number was made, and checking one step at a time (rather than judging from the final answer alone) is what locates it.
Full model answer. (a)
The error is in step 3. Checking each step: step 1 divides 130.8 g by Zn's own molar mass, 65.4 g/mol — correct. Step 2 reads the ratio coefficient of ZnS ÷ coefficient of Zn = 1 ÷ 1 from the balanced equation — correct. Step 3 must use the WANTED substance's molar mass, ZnS's 97.5 g/mol, but the student used Zn's 65.4 g/mol — the given substance's molar mass. That is why the answer came out equal to the starting mass.
(b)
Corrected step 3:
moles of ZnS = 2.0 mol
m = n × M
m of ZnS = 2.0 × 97.5
m of ZnS = 195.0 g
(c)
The wrong number was made in step 3 itself: steps 1 and 2 both made correct numbers, so the classmate's claim that the calculation is wrong from step 1 onward fails. No step in this calculation merely inherited a wrong number — nothing comes after step 3. The error lives in step 3, where the wrong number was made, and checking one step at a time (rather than judging from the final answer alone) is what locates it.
End of Topic Test: Limiting reactant (u08-t2 · 5 forms × 22 items · 110 total)
David's notes for James:
CALC-HEAVY FRQ SPLIT - frq_per_form = 2, OVERRIDING your gate data's frq_per_form=1 baseline (confirm the composition): QB-v2 sec.4 directs 2-3 FRQs for calculation-heavy / multi-step-procedure topics and names stoichiometry outright, and this scope IS the limiting-reactant procedure. Count = one FRQ per DISTINCT taught multi-step procedure: FRQ-1 at seq 21 = STO-020's moles-given needed-versus-available routine (medium/application); FRQ-2 at seq 22 = STO-022's masses-given 4-step capstone chain (hard/application; n=m/M both reactants -> limiting check -> mole ratio -> m=nxM). A THIRD FRQ was considered and REJECTED: the only remaining production surface is STO-017's draw-the-after-box skill, which is already measured by two objective image-option rows (seq 8/15, picture distractors carrying the same misconception classes) and cannot be typed without degrading to a counting list - the standing drawing-FRQ typed-plan question below applies. Total 20 objective + 2 FRQ = 22 items/form (inside the ~21-23 calc-heavy window); D5 scaling NOT triggered (9 lessons -> min(20, 54) = 20; your own target_size field says 20 and agrees).
Exact-ratio 'Neither reactant' COURSE-AUTHORITY ruling (confirm): 'Neither reactant' / 'Neither - the two reactants run out together' appears as keys and distractors across the particle-diagram and moles-routine rows. These are NOT none-of-the-above constructions (V-T5 PASS; zero NOTA/AOTA replacements were needed this gate - your seed pool carried none): the course explicitly teaches the exact-ratio outcome ('The two supplies run out together, so neither reactant is limiting and nothing is left over' - STO-016 walkthrough; your own STO-020 gate items key 'Neither - the two reactants run out together'). One blind reviewer flagged the convention repeatedly under general-chemistry framing; STO-015's 'exactly one of each role' rule is self-scoped ('Every reaction that stops THIS WAY...'), so course-local definitions carry. Key picks were 214/214 across all blind rounds regardless.
Molar-mass precision convention (confirm): the course's 1-decimal molar masses are treated as EXACT constants (your STO-022 worked examples do the same), and after sol QC every quantitative item now STATES it in-stem: 'Molar masses (exact): ...'. Limiting-chain intermediates are displayed at supporting precision (e.g. 0.5000 / 0.2000 mol) in numeric keyExplanations and FRQ model answers; FRQ rubrics carry '(or a correctly rounded equivalent such as ...)' and '(any equivalent shown precision)' tolerances so a correctly-rounded student answer cannot lose credit. Related recorded adjudication: B#22 (1.4 g H2) and D#22 (1.6 g) FRQ givens stay 2-sf because they are comparison-only givens whose model answers report the exact chain result - the taught 5-line scaffold treats givens as exact; givens an ANSWER flows through were padded to answer precision instead (A#19 29.20 g, E#19 5.400 g, C#22 49.80 g, E#22 0.8000 g). Mole givens like '8.0 mol' remain course-style 1-decimal.
Seed provenance (confirm the dispositions): pool 27 = 25 objective gateItems + 2 STO-017 gate productions; demo_curated 3 are all pool duplicates (pool[24]/[25]/[26], cited under both refs). 17 of the 20 form-A objective slots carry your seeds (forms B-E authored fresh; both FRQs are our authored parallels of the taught STO-020/STO-022 procedures). Eight placements VERBATIM (pool[4], [7], [13], [14], [16], [22] tail-rephrased only, [26], plus [8] reaction-freshened family). Interaction CONVERSIONS (U01-G3 precedent, confirm): pool[10] STO-017 production -> image-option MCQ at A#15; pool[23] STO-021 error-analysis MCQ -> numeric ENTRY at A#18 (the HS numeric-entry house rule for quant lessons). SEED RULING freshenings (instances only; structures, key logic, misconception classes verbatim): pool[0] liquid-oxygen/oxygen-tank -> liquid-oxidizer/oxidizer-tank; pool[6] 2Mg+O2 (corpus text) -> 2Cu+S; pool[8] CH4+2O2 -> 2Zn+O2; pool[11] N2+3H2 (corpus) -> fresh after-box reaction; pool[13] opening 'Carbon burns in a sealed jar' (duplicates STO-015 WE2) -> 'Lumps of carbon burn inside a sealed jar'; pool[15] sodium/oxygen name-words dropped to formulas; pool[18] CH4+2O2 -> steam-iron reaction; pool[19] 2H2+O2 -> ethane combustion; pool[24] molar masses burned -> reaction freshened AND asked product moved to FeCl2 (any 'mass of H2' run collides with STO-022 WE1's H2O token stream).
Seed adjacency adjudication (recorded): with 17 of 20 objective rows seeded, pairwise non-adjacency of seeded rows is arithmetically impossible - instance-level separation is enforced instead (every reaction equation lives at exactly one seq; material words are seq-laned; every non-trivial number unique across all five forms and off your spec's burned-number list; 54 unique scanned numbers in the final package). G1 row-order-tweak precedent extended.
V-T9 x specNotes format adjudication (recorded): your STO-020/021/022 mcq seed explanations carry scaffold stage-marker labels ('Write down the equation', 'Substitute in the values') which our scaffold-boundary validator forbids on non-numeric items. Placed seed mcq explanations keep the full computation and compare line with the marker LABELS dropped; numeric items keep the full taught scaffold; FRQ model answers show the full working without marker labels (U01-G3 FRQ canon style). Content preserved, format adjusted - the shared validator was NOT changed.
STO-020 rows are MCQ by your own specNotes ('categorical answer means mcq4 with bare substance options (R11) while every feedback chain/explanation runs the full scaffold') - the moles-routine numeric WORK is exercised in FRQ-1's parts instead. Numeric ENTRY rows sit at seq 16/18 (STO-021 product moles) and seq 19 (STO-022 full masses chain), per the HS numeric-entry rule for quant lessons.
Accepted advisory classes from blind review (recorded, no fix): (a) cross-item cueing - definitional recall rows state principles later calc rows use; inherent to the easy->hard recall->application blueprint; (b) definition-echo recall keys; (c) longest-option flags where the deterministic V-T4 longest-key balance PASSES; (d) product-as-limiting distractors - the taught misconception class. Three sol minors survive as recorded style notes (B#5 stem-contradicted-distractor style; E#19/E#22 non-limiting intermediates shown at 0.40/0.50 rather than padded 4-dp - numerically exact, keys unaffected).
Instance hygiene: classroom-safe chemistry throughout - zero fissile/weapons/dose framings bank-wide (re-swept at packaging, 0 hits). Course-corpus-burned reactions (2CO+O2, N2+3H2, 2H2+O2, CH4+2O2, 2Mg+O2, Zn+S, 2Al+3Cl2, Zn+2HCl, 4Fe+3O2, 2Na+Cl2, H2+Cl2, 2SO2+O2) appear in NO test item. One cross-lesson-material WARN pair stands adjudicated (A#9 carbon/oxygen - your own gate item; C#5 vinegar), screened for memory-answerability at blind review (keys picked without lesson access).
CRASH-RESUME PROVENANCE (disclosure, no action needed): the authoring fleet was killed twice mid-run by laptop restarts (after stage 2, and again during packaging). Both resumes worked from DISK STATE with full audits recorded in fix_log.md: every gate JSON artifact json.loads-verified (38/38 clean, nothing truncated), forms re-verified byte-identical to their validated versions, and the deterministic validator re-run fresh each time (12/12 PASS, exit 0, byte-identical output to the pre-crash runs). QC itself ran uninterrupted end-to-end after the first resume: blind rounds r1-r10 (key picks 214/214, keyless 66/66 graded match), sol authoritative loops 1-4 to 0 FAILs (110/110 pass). This export was built from the crash-audited forms_final.json.
FRQ delivery: your `prod` draw-on-paper class is delivered as a typed diagram-PLAN task (state kinds/recipes/counts/spacing; rubric grades the same observables). Confirm this mapping (pending from the U01-G1 round-trip).
full notes file
# u08-t2 — notes/queries from David's side (U8-G2)
1. CALC-HEAVY FRQ SPLIT - frq_per_form = 2, OVERRIDING your gate data's frq_per_form=1 baseline (confirm the composition): QB-v2 sec.4 directs 2-3 FRQs for calculation-heavy / multi-step-procedure topics and names stoichiometry outright, and this scope IS the limiting-reactant procedure. Count = one FRQ per DISTINCT taught multi-step procedure: FRQ-1 at seq 21 = STO-020's moles-given needed-versus-available routine (medium/application); FRQ-2 at seq 22 = STO-022's masses-given 4-step capstone chain (hard/application; n=m/M both reactants -> limiting check -> mole ratio -> m=nxM). A THIRD FRQ was considered and REJECTED: the only remaining production surface is STO-017's draw-the-after-box skill, which is already measured by two objective image-option rows (seq 8/15, picture distractors carrying the same misconception classes) and cannot be typed without degrading to a counting list - the standing drawing-FRQ typed-plan question below applies. Total 20 objective + 2 FRQ = 22 items/form (inside the ~21-23 calc-heavy window); D5 scaling NOT triggered (9 lessons -> min(20, 54) = 20; your own target_size field says 20 and agrees).
2. Exact-ratio 'Neither reactant' COURSE-AUTHORITY ruling (confirm): 'Neither reactant' / 'Neither - the two reactants run out together' appears as keys and distractors across the particle-diagram and moles-routine rows. These are NOT none-of-the-above constructions (V-T5 PASS; zero NOTA/AOTA replacements were needed this gate - your seed pool carried none): the course explicitly teaches the exact-ratio outcome ('The two supplies run out together, so neither reactant is limiting and nothing is left over' - STO-016 walkthrough; your own STO-020 gate items key 'Neither - the two reactants run out together'). One blind reviewer flagged the convention repeatedly under general-chemistry framing; STO-015's 'exactly one of each role' rule is self-scoped ('Every reaction that stops THIS WAY...'), so course-local definitions carry. Key picks were 214/214 across all blind rounds regardless.
3. Molar-mass precision convention (confirm): the course's 1-decimal molar masses are treated as EXACT constants (your STO-022 worked examples do the same), and after sol QC every quantitative item now STATES it in-stem: 'Molar masses (exact): ...'. Limiting-chain intermediates are displayed at supporting precision (e.g. 0.5000 / 0.2000 mol) in numeric keyExplanations and FRQ model answers; FRQ rubrics carry '(or a correctly rounded equivalent such as ...)' and '(any equivalent shown precision)' tolerances so a correctly-rounded student answer cannot lose credit. Related recorded adjudication: B#22 (1.4 g H2) and D#22 (1.6 g) FRQ givens stay 2-sf because they are comparison-only givens whose model answers report the exact chain result - the taught 5-line scaffold treats givens as exact; givens an ANSWER flows through were padded to answer precision instead (A#19 29.20 g, E#19 5.400 g, C#22 49.80 g, E#22 0.8000 g). Mole givens like '8.0 mol' remain course-style 1-decimal.
4. Seed provenance (confirm the dispositions): pool 27 = 25 objective gateItems + 2 STO-017 gate productions; demo_curated 3 are all pool duplicates (pool[24]/[25]/[26], cited under both refs). 17 of the 20 form-A objective slots carry your seeds (forms B-E authored fresh; both FRQs are our authored parallels of the taught STO-020/STO-022 procedures). Eight placements VERBATIM (pool[4], [7], [13], [14], [16], [22] tail-rephrased only, [26], plus [8] reaction-freshened family). Interaction CONVERSIONS (U01-G3 precedent, confirm): pool[10] STO-017 production -> image-option MCQ at A#15; pool[23] STO-021 error-analysis MCQ -> numeric ENTRY at A#18 (the HS numeric-entry house rule for quant lessons). SEED RULING freshenings (instances only; structures, key logic, misconception classes verbatim): pool[0] liquid-oxygen/oxygen-tank -> liquid-oxidizer/oxidizer-tank; pool[6] 2Mg+O2 (corpus text) -> 2Cu+S; pool[8] CH4+2O2 -> 2Zn+O2; pool[11] N2+3H2 (corpus) -> fresh after-box reaction; pool[13] opening 'Carbon burns in a sealed jar' (duplicates STO-015 WE2) -> 'Lumps of carbon burn inside a sealed jar'; pool[15] sodium/oxygen name-words dropped to formulas; pool[18] CH4+2O2 -> steam-iron reaction; pool[19] 2H2+O2 -> ethane combustion; pool[24] molar masses burned -> reaction freshened AND asked product moved to FeCl2 (any 'mass of H2' run collides with STO-022 WE1's H2O token stream).
5. Seed adjacency adjudication (recorded): with 17 of 20 objective rows seeded, pairwise non-adjacency of seeded rows is arithmetically impossible - instance-level separation is enforced instead (every reaction equation lives at exactly one seq; material words are seq-laned; every non-trivial number unique across all five forms and off your spec's burned-number list; 54 unique scanned numbers in the final package). G1 row-order-tweak precedent extended.
6. V-T9 x specNotes format adjudication (recorded): your STO-020/021/022 mcq seed explanations carry scaffold stage-marker labels ('Write down the equation', 'Substitute in the values') which our scaffold-boundary validator forbids on non-numeric items. Placed seed mcq explanations keep the full computation and compare line with the marker LABELS dropped; numeric items keep the full taught scaffold; FRQ model answers show the full working without marker labels (U01-G3 FRQ canon style). Content preserved, format adjusted - the shared validator was NOT changed.
7. STO-020 rows are MCQ by your own specNotes ('categorical answer means mcq4 with bare substance options (R11) while every feedback chain/explanation runs the full scaffold') - the moles-routine numeric WORK is exercised in FRQ-1's parts instead. Numeric ENTRY rows sit at seq 16/18 (STO-021 product moles) and seq 19 (STO-022 full masses chain), per the HS numeric-entry rule for quant lessons.
8. Accepted advisory classes from blind review (recorded, no fix): (a) cross-item cueing - definitional recall rows state principles later calc rows use; inherent to the easy->hard recall->application blueprint; (b) definition-echo recall keys; (c) longest-option flags where the deterministic V-T4 longest-key balance PASSES; (d) product-as-limiting distractors - the taught misconception class. Three sol minors survive as recorded style notes (B#5 stem-contradicted-distractor style; E#19/E#22 non-limiting intermediates shown at 0.40/0.50 rather than padded 4-dp - numerically exact, keys unaffected).
9. Instance hygiene: classroom-safe chemistry throughout - zero fissile/weapons/dose framings bank-wide (re-swept at packaging, 0 hits). Course-corpus-burned reactions (2CO+O2, N2+3H2, 2H2+O2, CH4+2O2, 2Mg+O2, Zn+S, 2Al+3Cl2, Zn+2HCl, 4Fe+3O2, 2Na+Cl2, H2+Cl2, 2SO2+O2) appear in NO test item. One cross-lesson-material WARN pair stands adjudicated (A#9 carbon/oxygen - your own gate item; C#5 vinegar), screened for memory-answerability at blind review (keys picked without lesson access).
10. CRASH-RESUME PROVENANCE (disclosure, no action needed): the authoring fleet was killed twice mid-run by laptop restarts (after stage 2, and again during packaging). Both resumes worked from DISK STATE with full audits recorded in fix_log.md: every gate JSON artifact json.loads-verified (38/38 clean, nothing truncated), forms re-verified byte-identical to their validated versions, and the deterministic validator re-run fresh each time (12/12 PASS, exit 0, byte-identical output to the pre-crash runs). QC itself ran uninterrupted end-to-end after the first resume: blind rounds r1-r10 (key picks 214/214, keyless 66/66 graded match), sol authoritative loops 1-4 to 0 FAILs (110/110 pass). This export was built from the crash-audited forms_final.json.
To decide which reactant is limiting, what does the needed-versus-available comparison actually compare?
The moles of one reactant needed to use up the other, against the moles actually available. KEY
The raw moles of one reactant measured against the raw moles of the other, with no scaling.
The coefficients of the two reactants, on their own.
The masses of the two reactants in grams.
STO-015 · easy · application · seq 4 · seed:U8-G2 pool[4] (STO-015, James) — VERBATIM; phosphorus appears in no scope corpus field · mcq4
In a certain reaction, phosphorus is the excess reactant. What must be true when the reaction stops?
Some phosphorus is left over, unreacted. KEY
The phosphorus was completely used up first.
The phosphorus never reacted at all.
The phosphorus turned into the limiting reactant as the amounts fell.
STO-014 · medium · application · seq 5 · seed:U8-G2 pool[0] (STO-014, James) — freshened: 'liquid oxygen'->'liquid oxidizer' and 'oxygen tank'->'oxidizer tank' ('oxygen' is an own-lesson corpus material); all other fields verbatim, swept for the old word (QB 3.9) · mcq4
A rocket engine burns fuel and liquid oxidizer fed from two separate tanks. Mid-flight, the engine shuts down smoothly at its normal running temperature, even though the fuel tank still holds fuel. Which explanation best fits?
The oxidizer tank ran empty, so the burning reaction stopped. KEY
The fuel ran out at the same moment the engine stopped.
The engine got too hot for the reaction to continue.
Burning reactions can only run for a fixed number of minutes.
STO-016 · medium · application · seq 6 · seed:U8-G2 pool[6] (STO-016, James) — must_freshen_instance: reaction 2Mg+O₂→2MgO is corpus text; freshened to 2Cu + S → Cu₂S at the same equal-counts geometry (4 v 4, 2:1 demand); structure, bare labels, and all three misconception classes kept · mcq4
The figure shows a reactant mixture for the reaction 2Cu + S → Cu₂S. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "four Cu atoms (each one medium gray circle, unjoined) and four S atoms (each one small yellow circle, unjoined), mixed together, widely spaced, no Cu touchi… · key: "Cu = one medium gray circle · S = one small yellow circle" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four Cu atoms (each one medium gray circle, unjoined) and four S atoms (each one small yellow circle, unjoined), mixed together, widely spaced, no Cu touching an S",
"key": "Cu = one medium gray circle · S = one small yellow circle",
"style": "clean textbook particle diagram, single labeled box"
}
}
The figure shows a reactant mixture for the reaction 2NO + O₂ → 2NO₂. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "four NO molecules (each one medium blue circle joined to one large red circle) and two O₂ molecules (each two large red circles joined), mixed together, wid… · key: "NO = one medium blue circle joined to one large red circle · O₂ = two large red circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four NO molecules (each one medium blue circle joined to one large red circle) and two O₂ molecules (each two large red circles joined), mixed together, widely spaced, no two different molecules touching",
"key": "NO = one medium blue circle joined to one large red circle · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
Neither reactant KEY
NO
O₂
NO₂
STO-017 · medium · application · seq 8 · seed:U8-G2 pool[11] (STO-017, James) — must_freshen_instance: reaction N₂+3H₂→2NH₃ is corpus text (specNotes flag the overlap); freshened to Zn + I₂ → ZnI₂ (3 Zn + 5 I₂) with the same leftover-excess geometry; the three picture-distractor classes kept verbatim in sense · mcq4
The figure shows a reactant mixture. It reacts by Zn + I₂ → ZnI₂. Which particle diagram shows the mixture after the reaction stops?
FIGURE (stimulus · stimulus) — see raw specraw spec
{
"option_format": "image",
"render_contract": "R2 image-XOR-text: options render as pictures ONLY; the 'text'/'key' placeholder strings are internal ids for validation and are never rendered and never used for aria (R23 — descriptions must not exceed what a sighted student sees).",
"figureRequired": true,
"figureShape": "stimulus",
"stimulus": {
"figureType": "particle-diagram",
"box": "three Zn atoms (each one medium gray circle, unjoined) and five I₂ molecules (each two large purple circles joined), mixed together, widely spaced, no Zn touching an I₂",
"key": "Zn = one medium gray circle · I₂ = two large purple circles joined",
"style": "clean textbook particle diagram, single labeled box"
},
"optionFigureSpecs": [
{
"option": "key",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three ZnI₂ units (each one medium gray circle joined to two large purple circles) and two I₂ molecules (each two large purple circles joined), mixed together, widely spaced",
"key": "ZnI₂ = one medium gray circle joined to two large purple circles · I₂ = two large purple circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor1",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two ZnI₂ units, one lone Zn atom (one medium gray circle), and three I₂ molecules, mixed together, widely spaced",
"key": "ZnI₂ = one medium gray circle joined to two large purple circles · Zn = one medium gray circle · I₂ = two large purple circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor2",
"figureSpec": {
"figureType": "particle-diagram",
"box": "five ZnI₂ units (each one medium gray circle joined to two large purple circles), no other particles, mixed together, widely spaced",
"key": "ZnI₂ = one medium gray circle joined to two large purple circles",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor3",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three ZnI₂ units (each one medium gray circle joined to two large purple circles) only, no other particles, mixed together, widely spaced",
"key": "ZnI₂ = one medium gray circle joined to two large purple circles",
"style": "clean textbook particle diagram, single labeled box"
}
}
]
}
Picture 1 KEY
Picture 2
Picture 3
Picture 4
STO-018 · medium · application · seq 9 · seed:U8-G2 pool[13] (STO-018, James) — opening tokens freshened ('Carbon burns in a sealed jar' duplicates STO-015 WE2 text; SEED RULING); options, key, and explanation verbatim · mcq4
Lumps of carbon burn inside a sealed jar holding a small amount of oxygen, forming CO₂ until the reaction stops, with unburned carbon left over. Which single change would let MORE CO₂ form?
Adding more oxygen to the jar. KEY
Adding more carbon to the jar.
Shaking the jar to mix the leftover carbon better.
Sealing the jar more tightly so nothing escapes.
STO-019 · medium · application · seq 10 · seed:U8-G2 pool[15] (STO-019, James) — formula_only_restyle: stem's 'sodium'/'oxygen' name-words dropped ('oxygen' is an own-lesson corpus material); reasoning options and explanation otherwise verbatim, swept · mcq4
6.0 mol of Na and 2.0 mol of O₂ react by 4Na + O₂ → 2Na₂O. Which reactant is the limiting reactant, and why?
Na — using all 2.0 mol of O₂ would need 8.0 mol of Na, and only 6.0 mol is available. KEY
O₂ — there is much less O₂ than Na, and the reactant present in the smaller amount runs out first.
Na — metals always run out first because they react fastest.
Neither — 6.0 mol of Na is more than enough for 2.0 mol of O₂.
STO-020 · medium · application · seq 11 · seed:U8-G2 pool[19] (STO-020, James) — must_freshen_instance: 2H₂+O₂→2H₂O is corpus text; freshened to ethane combustion at the same branch geometry (bigger pile limiting); bare labels kept; explanation V-T9-restyled (computation + compare line, no stage-marker labels) · mcq4
2.0 mol of C₂H₆ and 5.0 mol of O₂ react by 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. Which reactant is the limiting reactant?
O₂ KEY
C₂H₆
CO₂
Neither reactant
STO-020 · medium · application · seq 12 · seed:U8-G2 pool[18] (STO-020, James) — must_freshen_instance: CH₄+2O₂ is corpus text; freshened to the steam-iron reaction at the same exact-ratio geometry; option set and misconception classes kept; V-T9 restyle · mcq4
3.0 mol of Fe and 4.0 mol of H₂O (as steam) react by 3Fe + 4H₂O → Fe₃O₄ + 4H₂. Which statement about the limiting reactant is correct?
Neither — the two reactants run out together. KEY
Fe is the limiting reactant.
H₂O is the limiting reactant.
It cannot be decided from the moles alone — the masses of the two reactants are needed as well.
STO-019 · medium · transfer · seq 13 · seed:U8-G2 pool[16] (STO-019, James) — VERBATIM (instance-free stem; the N₂/H₂ counterexample lives in the explanation, which V-T7 does not scan) · mcq4
After one homework problem where the smaller amount was the limiting reactant, a student concludes: 'The reactant you have less of is always the limiting reactant.' Which reply is correct?
Not always — the coefficients decide how much of each reactant is needed, so a reactant can be limiting even when there is more of it. KEY
Correct — the reactant supplied in the smaller amount is the one that runs out first, whatever the equation says.
Not always — the limiting reactant is actually the reactant with the larger coefficient in the balanced chemical equation.
The shortcut holds once both amounts are converted to moles; only gram amounts can break it.
STO-018 · medium · transfer · seq 14 · seed:U8-G2 pool[14] (STO-018, James) — VERBATIM (instance-free) · mcq4
A student claims: 'Even after one reactant runs out, a little product keeps forming slowly from the reactant that is left.' Which reply is correct?
The student is wrong — after one reactant runs out, no product forms at all. KEY
The student is right — the leftover reactant keeps reacting with itself, just more slowly.
The student is right, as long as the mixture stays warm enough.
No more product forms, but only because the leftover reactant is chemically ruined — it could not react even if a fresh supply of the used-up reactant were added.
STO-017 · medium · application · seq 15 · seed:U8-G2 pool[10] (STO-017 gate PRODUCTION, James) — interaction_conversion production->image-mcq (U01-G3 precedent): the model-answer box is the key picture, checklist violations are the distractor pictures; reaction and counts (3 S + 3 O₂) kept verbatim (gate items are not corpus) · mcq4
The figure shows a reactant mixture for the reaction S + O₂ → SO₂. Which particle diagram shows the mixture after the reaction stops?
FIGURE (stimulus · stimulus) — see raw specraw spec
{
"option_format": "image",
"render_contract": "R2 image-XOR-text: options render as pictures ONLY; the 'text'/'key' placeholder strings are internal ids for validation and are never rendered and never used for aria (R23 — descriptions must not exceed what a sighted student sees).",
"figureRequired": true,
"figureShape": "stimulus",
"stimulus": {
"figureType": "particle-diagram",
"box": "three S atoms (each one small yellow circle, unjoined) and three O₂ molecules (each two large red circles joined), mixed together, widely spaced, no S touching an O₂",
"key": "S = one small yellow circle · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
},
"optionFigureSpecs": [
{
"option": "key",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three SO₂ molecules (each one small yellow circle joined to two large red circles), mixed together, widely spaced, no other particles",
"key": "SO₂ = one small yellow circle joined to two large red circles",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor1",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two SO₂ molecules, one S atom (one small yellow circle), and one O₂ molecule (two large red circles joined), mixed together, widely spaced",
"key": "SO₂ = one small yellow circle joined to two large red circles · S = one small yellow circle · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor2",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three SO₂ molecules (each one small yellow circle joined to two large red circles) and one O₂ molecule (two large red circles joined), mixed together, widely spaced",
"key": "SO₂ = one small yellow circle joined to two large red circles · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor3",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three product units drawn with the wrong recipe: each one small yellow circle joined to three large red circles, no other particles, mixed together, widely spaced",
"key": "product drawn as one small yellow circle joined to three large red circles",
"style": "clean textbook particle diagram, single labeled box"
}
}
]
}
Picture 1 KEY
Picture 2
Picture 3
Picture 4
STO-021 · medium · application · seq 16 · seed:U8-G2 pool[22] (STO-021, James) — question tail freshened ('can form? Give your answer' duplicates STO-021 fluency stems; SEED RULING); numbers, equation, scaffold, and misconceptions verbatim · numeric
Propane burns by C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. O₂ is the limiting reactant with 10.0 mol available, and 4.0 mol of C₃H₈ is present. How many moles of CO₂ form from this mixture? Give your answer in moles to 2 significant figures.
Answer: 6.0 mol KEY
also accepted: 6.0 mol
wrongValueDiagnostics
[
{
"value": 10.0,
"misconception": "you carried the limiting reactant's moles straight across without the mole ratio",
"feedback": null
},
{
"value": 12.0,
"misconception": "you started from the excess reactant C₃H₈ — the limiting reactant O₂ caps the product",
"feedback": null
},
{
"value": 16.7,
"misconception": "you flipped the mole ratio — the coefficient of CO₂ goes on top and O₂'s underneath",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of O₂ (limiting) = 10.0 mol\nThe 4.0 mol of C₃H₈ is the excess reactant — it does not enter the calculation.\nWrite down the equation:\nmoles of CO₂ = moles of O₂ × (coefficient of CO₂ ÷ coefficient of O₂)\nSubstitute in the values, and calculate:\nmoles of CO₂ = 10.0 × (3 ÷ 5)\nmoles of CO₂ = 6.0 mol"
}
STO-016 · medium · transfer · seq 17 · seed:U8-G2 pool[8] (STO-016, James) — must_freshen_instance: reaction CH₄+2O₂ is corpus text; freshened to 2Zn + O₂ → 2ZnO at a bigger-pile-limiting geometry; bare labels and misconception classes kept · mcq4
The figure shows a reactant mixture for the reaction 2Zn + O₂ → 2ZnO. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "six Zn atoms (each one medium gray circle, unjoined) and four O₂ molecules (each two large red circles joined), mixed together, widely spaced, no Zn touchin… · key: "Zn = one medium gray circle · O₂ = two large red circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "six Zn atoms (each one medium gray circle, unjoined) and four O₂ molecules (each two large red circles joined), mixed together, widely spaced, no Zn touching an O₂",
"key": "Zn = one medium gray circle · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
Zn KEY
O₂
ZnO
Neither reactant
STO-021 · hard · application · seq 18 · seed:U8-G2 pool[23] (STO-021, James) — interaction_conversion mcq->numeric ENTRY (U01-G3 precedent): the error-analysis narrative becomes the misconceptions list (started-from-excess primary); stem restyled formula-only ('copper'/'oxygen' words dropped); mechanism sentence + scaffold kept · numeric
In 2Cu + O₂ → 2CuO, O₂ is the limiting reactant with 1.0 mol available; the vessel also holds 6.0 mol of Cu. How many moles of CuO does the reaction produce? Give your answer in moles to 2 significant figures.
Answer: 2.0 mol KEY
also accepted: 2.0 mol
wrongValueDiagnostics
[
{
"value": 6.0,
"misconception": "you started from the excess reactant Cu — once the 1.0 mol of O₂ is used up, no more CuO can form no matter how much Cu remains",
"feedback": null
},
{
"value": 12.0,
"misconception": "you multiplied the excess reactant's moles by the ratio — the calculation must start from the limiting reactant O₂",
"feedback": null
},
{
"value": 0.5,
"misconception": "you flipped the mole ratio — the coefficient of CuO goes on top and O₂'s underneath",
"feedback": null
}
]
feedback
{
"keyExplanation": "Product can only form while every reactant is still available, so the reactant that runs out first stops product formation. O₂ is limiting, so the calculation starts from its 1.0 mol.\nWrite down the values in the question:\nmoles of O₂ (limiting) = 1.0 mol\nThe 6.0 mol of Cu is the excess reactant — it does not enter the calculation, and most of it is left over when the reaction stops.\nWrite down the equation:\nmoles of CuO = moles of O₂ × (coefficient of CuO ÷ coefficient of O₂)\nSubstitute in the values, and calculate:\nmoles of CuO = 1.0 × (2 ÷ 1)\nmoles of CuO = 2.0 mol"
}
STO-022 · hard · application · seq 19 · seed:U8-G2 pool[24] = demo_curated[0] (STO-022, James) — must_freshen_instance: reaction freshened (burned molar masses) AND the asked product moved to FeCl₂ ('mass of H₂ ...' token runs collide with STO-022 WE1's H₂O text); 4-step structure and misconception classes kept; molar-mass list + question phrasing restyled off the fluency-stem idiom (SEED RULING) · numeric
27.9 g of Fe and 29.20 g of HCl react by Fe + 2HCl → FeCl₂ + H₂. Molar masses (exact): Fe 55.8 g/mol, HCl 36.5 g/mol, FeCl₂ 126.8 g/mol. Give the mass of FeCl₂ that forms, in grams (4 significant figures).
Answer: 50.72 g KEY
also accepted: 50.72 g
wrongValueDiagnostics
[
{
"value": 63.4,
"misconception": "you calculated from the excess reactant Fe — using up 0.50 mol of Fe would need 1.0 mol of HCl, and only 0.8000 mol is available, so HCl is limiting",
"feedback": null
},
{
"value": 57.1,
"misconception": "you added the two reactant masses — the leftover excess iron is not part of the product",
"feedback": null
},
{
"value": 101.44,
"misconception": "you skipped the mole-ratio step — 0.80 mol is the HCl's moles, and every 2 mol of HCl makes only 1 mol of FeCl₂",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Fe = 27.9 g, M of Fe = 55.8 g/mol, m of HCl = 29.20 g, M of HCl = 36.5 g/mol, M of FeCl₂ = 126.8 g/mol\nStep 1 — convert each mass to moles:\nn(Fe) = 27.9 / 55.8 = 0.50 mol\nn(HCl) = 29.20 / 36.5 = 0.8000 mol\nStep 2 — identify the limiting reactant:\nmoles of HCl needed = 0.50 × (2 ÷ 1) = 1.0 mol — only 0.8000 mol is available, so HCl is limiting\nStep 3 — moles of product from the limiting reactant:\nmoles of FeCl₂ = 0.8000 × (1 ÷ 2) = 0.4000 mol\nStep 4 — convert the product moles to mass:\nm(FeCl₂) = 0.4000 × 126.8\nm(FeCl₂) = 50.72 g"
}
STO-022 · hard · application · seq 20 · seed:U8-G2 pool[26] = demo_curated[2] (STO-022, James) — VERBATIM (instance-free; option order differs between the two source copies, notational only in our key+distractors schema) · mcq4
A question gives the masses of both reactants and asks for the mass of product. Which plan gives the correct answer?
Convert both masses to moles, find the limiting reactant, take the product moles from it, then convert to mass. KEY
Add the two reactant masses together and convert the combined total straight into the product mass.
Calculate a product mass from each reactant separately and report the larger of the two answers.
Convert the smaller mass to moles and calculate the product straight from it, skipping any comparison.
STO-020 · medium · application · seq 21 · authored (FRQ-1 form A: needed > available branch; vessel framing avoids the STO-019 'are mixed' fluency idiom; model answer shows the working without stage-marker labels — V-T9 non-numeric boundary, U01-G3 FRQ canon) · frq
A sealed vessel holds 8.0 mol of P and 5.0 mol of O₂. The reaction is 4P + 5O₂ → 2P₂O₅.
A sealed vessel holds 8.0 mol of P and 5.0 mol of O₂. The reaction is 4P + 5O₂ → 2P₂O₅.
(a) Calculate how many moles of O₂ would be needed to react completely with all 8.0 mol of P. Show your working.
(b) Compare the moles of O₂ needed with the moles of O₂ available, and say which reactant is limiting.
(c) Name the excess reactant, and state what will be true of it when the reaction stops.
Part a. Calculate how many moles of O₂ would be needed to react completely with all 8.0 mol of P. Show your working.
Model: Amounts supplied: 8.0 mol of phosphorus, 5.0 mol of oxygen.
moles of oxygen needed = moles of phosphorus × (coefficient of oxygen ÷ coefficient of phosphorus)
moles of oxygen needed = 8.0 × (5 ÷ 4)
moles of oxygen needed = 10.0 mol
Part b. Compare the moles of O₂ needed with the moles of O₂ available, and say which reactant is limiting.
Model: 10.0 mol needed > 5.0 mol available — less oxygen is available than needed, so the oxygen runs out first. The oxygen is the limiting reactant.
Part c. Name the excess reactant, and state what will be true of it when the reaction stops.
Model: The excess reactant is the phosphorus. When the reaction stops, some of it will be left over, unreacted — it reacted only while the oxygen lasted.
Full model answer. (a) Amounts supplied: 8.0 mol of phosphorus, 5.0 mol of oxygen.
moles of oxygen needed = moles of phosphorus × (coefficient of oxygen ÷ coefficient of phosphorus)
moles of oxygen needed = 8.0 × (5 ÷ 4)
moles of oxygen needed = 10.0 mol
(b) 10.0 mol needed > 5.0 mol available — less oxygen is available than needed, so the oxygen runs out first. The oxygen is the limiting reactant.
(c) The excess reactant is the phosphorus. When the reaction stops, some of it will be left over, unreacted — it reacted only while the oxygen lasted.
STO-022 · hard · application · seq 22 · authored (FRQ-2 form A: acid limiting; larger-mole reactant NOT limiting; model answer de-labeled per V-T9 boundary) · frq
A student adds 35.61 g of Sn to a solution holding 32.36 g of HBr; the reaction is Sn + 2HBr → SnBr₂ + H₂.
A student adds 35.61 g of Sn to a solution holding 32.36 g of HBr; the reaction is Sn + 2HBr → SnBr₂ + H₂.
(a) Calculate the moles of each reactant.
(b) Identify the limiting reactant, showing the comparison of needed moles against available moles.
(c) Calculate the mass of SnBr₂ that forms. Show every step of your working.
Molar masses (exact): Sn 118.7 g/mol, HBr 80.9 g/mol, SnBr₂ 278.5 g/mol.
Part a. Calculate the moles of each reactant.
Model: moles of tin = 35.61 / 118.7 = 0.3000 mol
moles of hydrogen bromide = 32.36 / 80.9 = 0.4000 mol
Part b. Identify the limiting reactant, showing the comparison of needed moles against available moles.
Model: moles of hydrogen bromide needed = moles of tin × (coefficient ratio) = 0.60 mol
0.60 mol needed > 0.40 mol available — the hydrogen bromide falls short and runs out first: it is the limiting reactant.
Part c. Calculate the mass of SnBr₂ that forms. Show every step of your working.
Molar masses (exact): Sn 118.7 g/mol, HBr 80.9 g/mol, SnBr₂ 278.5 g/mol.
Model: moles of tin bromide = 0.4000 × (1 ÷ 2) = 0.2000 mol (starting from the limiting reactant)
mass of tin bromide = 0.2000 × 278.5
mass of tin bromide = 55.70 g
Full model answer. (a) moles of tin = 35.61 / 118.7 = 0.3000 mol
moles of hydrogen bromide = 32.36 / 80.9 = 0.4000 mol
(b) moles of hydrogen bromide needed = moles of tin × (coefficient ratio) = 0.60 mol
0.60 mol needed > 0.40 mol available — the hydrogen bromide falls short and runs out first: it is the limiting reactant.
(c) moles of tin bromide = 0.4000 × (1 ÷ 2) = 0.2000 mol (starting from the limiting reactant)
mass of tin bromide = 0.2000 × 278.5
mass of tin bromide = 55.70 g
Nickel powder reacts with a gas in a sealed flask. The reaction stops, and unreacted nickel powder is still visible at the bottom. Which label fits the nickel?
It is the excess reactant — it is the reactant left over. KEY
It is the limiting reactant — powders always run out before gases.
It is the product of the reaction.
It holds no role, because some of it never reacted.
STO-014 · medium · application · seq 5 · authored (fixed-supply class: sealed box; word lane: butane; O₂ written as formula only) · mcq4
A butane camping burner is lit inside a sealed demonstration box. After a while the flame dies, though the butane cartridge still reads half full. Why did the flame go out?
The box's sealed supply of O₂ ran out, starving the flame. KEY
The butane was completely used up.
The flame used up its heat and could not keep going.
The figure shows a reactant mixture for the reaction 2K + Br₂ → 2KBr. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "four K atoms (each one medium gray circle, unjoined) and four Br₂ molecules (each two large brown circles joined), mixed together, widely spaced, no K touch… · key: "K = one medium gray circle · Br₂ = two large brown circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four K atoms (each one medium gray circle, unjoined) and four Br₂ molecules (each two large brown circles joined), mixed together, widely spaced, no K touching a Br₂",
"key": "K = one medium gray circle · Br₂ = two large brown circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
K KEY
Br₂
KBr
Neither reactant
STO-016 · medium · application · seq 7 · authored (exact-ratio family; partial oxidation in limited O₂ — replaces CO+2H₂ whose CH₃OH token run collided with 11D) · mcq4
The figure shows a reactant mixture for the reaction 2C + O₂ → 2CO. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "four C atoms (each one medium black circle, unjoined) and two O₂ molecules (each two large red circles joined), mixed together, widely spaced, no C touching… · key: "C = one medium black circle · O₂ = two large red circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four C atoms (each one medium black circle, unjoined) and two O₂ molecules (each two large red circles joined), mixed together, widely spaced, no C touching an O₂",
"key": "C = one medium black circle · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
Neither reactant KEY
C
O₂
CO
STO-017 · medium · application · seq 8 · authored (leftover-excess family, excess = the smaller-molecule reactant) · mcq4
The figure shows a reactant mixture. It reacts by H₂ + Br₂ → 2HBr. Which particle diagram shows the mixture after the reaction stops?
FIGURE (stimulus · stimulus) — see raw specraw spec
{
"option_format": "image",
"render_contract": "R2 image-XOR-text: options render as pictures ONLY; the 'text'/'key' placeholder strings are internal ids for validation and are never rendered and never used for aria (R23 — descriptions must not exceed what a sighted student sees).",
"figureRequired": true,
"figureShape": "stimulus",
"stimulus": {
"figureType": "particle-diagram",
"box": "four H₂ molecules (each two small light circles joined) and two Br₂ molecules (each two large brown circles joined), mixed together, widely spaced, no two different molecules touching",
"key": "H₂ = two small light circles joined · Br₂ = two large brown circles joined",
"style": "clean textbook particle diagram, single labeled box"
},
"optionFigureSpecs": [
{
"option": "key",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four HBr molecules (each one small light circle joined to one large brown circle) and two H₂ molecules (each two small light circles joined), mixed together, widely spaced",
"key": "HBr = one small light circle joined to one large brown circle · H₂ = two small light circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor1",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two HBr molecules, one Br₂ molecule (two large brown circles joined), and three H₂ molecules, mixed together, widely spaced",
"key": "HBr = one small light circle joined to one large brown circle · Br₂ = two large brown circles joined · H₂ = two small light circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor2",
"figureSpec": {
"figureType": "particle-diagram",
"box": "eight HBr molecules (each one small light circle joined to one large brown circle), no other particles, mixed together, widely spaced",
"key": "HBr = one small light circle joined to one large brown circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor3",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four HBr molecules (each one small light circle joined to one large brown circle) only, no other particles, mixed together, widely spaced",
"key": "HBr = one small light circle joined to one large brown circle",
"style": "clean textbook particle diagram, single labeled box"
}
}
]
}
Picture 1 KEY
Picture 2
Picture 3
Picture 4
STO-018 · medium · application · seq 9 · authored (ask class: what-happens-if-more-excess-added; word lanes tin/iodine; spare pool[12]'s surface class with a fresh pair) · mcq4
A strip of tin reacts with iodine vapor in a sealed tube until the iodine is gone; leftover tin remains. The chemist wants more product. What happens if MORE TIN is pushed into the tube?
Nothing more forms — the iodine is gone, and the new tin has nothing to pair with. KEY
More product forms, because tin is a reactant and adding any reactant makes more product.
More product forms slowly, drawing on the iodine locked in the product.
The product splits up to share its iodine with the new tin.
STO-019 · medium · application · seq 10 · authored (shortcut-fails family: smaller amount NOT limiting) · mcq4
2.0 mol of P₄ and 8.0 mol of O₂ react by P₄ + 5O₂ → P₄O₁₀. Which reasoning correctly identifies the limiting reactant?
O₂ — using all 2.0 mol of P₄ would need 10.0 mol of O₂, and only 8.0 mol is available. KEY
P₄ — it is present in the smaller amount, and the smaller amount is always limiting.
O₂ — its coefficient is larger, so it is always the limiting reactant.
Neither — 8.0 mol of O₂ is clearly plenty for just 2.0 mol of P₄.
4.0 mol of NH₃ and 6.0 mol of O₂ react by 4NH₃ + 5O₂ → 4NO + 6H₂O. Which reactant is the limiting reactant?
NH₃ KEY
O₂
NO
Neither reactant
STO-020 · medium · application · seq 12 · authored (decimals case per specNotes; ledgered decimals 0.45/0.85 printed at this seq only) · mcq4
0.45 mol of Cr and 0.85 mol of Cl₂ react by 2Cr + 3Cl₂ → 2CrCl₃. Which reactant is the limiting reactant?
Cr KEY
Cl₂
CrCl₃
Neither reactant
STO-019 · medium · transfer · seq 13 · authored (claim class: coefficient-alone rule; instance-free) · mcq4
A lab partner offers a rule: 'The reactant with the larger coefficient in the balanced equation is always the limiting reactant, because it gets used up fastest.' Which reply is correct?
Not always — fast spending is not the same as running out; a big enough supply still outlasts the other reactant. KEY
Correct — the balanced equation alone always fixes which reactant runs out, whatever the supplied amounts are.
Correct for gases, but not for solids and liquids.
Not always — the limiting reactant is actually the one with the SMALLER coefficient.
STO-018 · medium · transfer · seq 14 · authored (claim class: leftover-turns-unreactive, with the restock-prediction twist) · mcq4
A classmate says: 'When its partner runs out, the leftover excess reactant turns unreactive — that is why the reaction stops.' Which reply is correct?
Wrong — the leftover stays fully reactive; the reaction stops only because its partner no longer exists in the mixture. KEY
Right — leftover reactants always lose their reactivity the moment a reaction comes to its end.
Right — that is why adding fresh partner reactant later does nothing.
Wrong — the leftover turns into product slowly by itself instead.
STO-017 · medium · application · seq 15 · authored (exact-ratio family; instance = spare gate production pool[9]'s reaction at exact-ratio counts — gate items are not corpus) · mcq4
The figure shows a reactant mixture for the reaction N₂ + O₂ → 2NO. Which particle diagram shows the mixture after the reaction stops?
FIGURE (stimulus · stimulus) — see raw specraw spec
{
"option_format": "image",
"render_contract": "R2 image-XOR-text: options render as pictures ONLY; the 'text'/'key' placeholder strings are internal ids for validation and are never rendered and never used for aria (R23 — descriptions must not exceed what a sighted student sees).",
"figureRequired": true,
"figureShape": "stimulus",
"stimulus": {
"figureType": "particle-diagram",
"box": "three N₂ molecules (each two medium blue circles joined) and three O₂ molecules (each two large red circles joined), mixed together, widely spaced, no two different molecules touching",
"key": "N₂ = two medium blue circles joined · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
},
"optionFigureSpecs": [
{
"option": "key",
"figureSpec": {
"figureType": "particle-diagram",
"box": "six NO molecules (each one medium blue circle joined to one large red circle), mixed together, widely spaced, no other particles",
"key": "NO = one medium blue circle joined to one large red circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor1",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four NO molecules, one N₂ molecule (two medium blue circles joined), and one O₂ molecule (two large red circles joined), mixed together, widely spaced",
"key": "NO = one medium blue circle joined to one large red circle · N₂ = two medium blue circles joined · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor2",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three NO molecules (each one medium blue circle joined to one large red circle), no other particles, mixed together, widely spaced",
"key": "NO = one medium blue circle joined to one large red circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor3",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three product units drawn with the wrong recipe: each one medium blue circle joined to two large red circles, no other particles, mixed together, widely spaced",
"key": "product drawn as one medium blue circle joined to two large red circles",
"style": "clean textbook particle diagram, single labeled box"
}
}
]
}
Picture 1 KEY
Picture 2
Picture 3
Picture 4
STO-021 · medium · application · seq 16 · authored (ratio-down case; blast-furnace instance — equation lives in Unit 8 Topic 1 lessons, outside this gate's scope corpus; 'iron' word removed, own-lesson material) · numeric
A blast furnace runs the reaction Fe₂O₃ + 3CO → 2Fe + 3CO₂. CO is the limiting reactant with 6.0 mol available, and 4.0 mol of Fe₂O₃ is present. How many moles of Fe form from this mixture? Give your answer in moles to 2 significant figures.
Answer: 4.0 mol KEY
also accepted: 4.0 mol
wrongValueDiagnostics
[
{
"value": 8.0,
"misconception": "you started from the excess reactant Fe₂O₃ — the limiting reactant CO caps the product",
"feedback": null
},
{
"value": 9.0,
"misconception": "you flipped the mole ratio — the coefficient of Fe goes on top and CO's underneath",
"feedback": null
},
{
"value": 6.0,
"misconception": "you carried the limiting reactant's moles straight across without the mole ratio",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of CO (limiting) = 6.0 mol\nThe 4.0 mol of Fe₂O₃ is the excess reactant — it does not enter the calculation.\nWrite down the equation:\nmoles of Fe = moles of CO × (coefficient of Fe ÷ coefficient of CO)\nSubstitute in the values, and calculate:\nmoles of Fe = 6.0 × (2 ÷ 3)\nmoles of Fe = 4.0 mol"
}
STO-016 · medium · transfer · seq 17 · authored (bigger-pile-limiting family, fresh reaction) · mcq4
The figure shows a reactant mixture for the reaction SiH₄ + 2O₂ → SiO₂ + 2H₂O. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "four SiH₄ molecules (each one medium dark circle joined to four small light circles) and six O₂ molecules (each two large red circles joined), mixed togethe… · key: "SiH₄ = one medium dark circle joined to four small light circles · O₂ = two large red circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four SiH₄ molecules (each one medium dark circle joined to four small light circles) and six O₂ molecules (each two large red circles joined), mixed together, widely spaced, no two different molecules touching",
"key": "SiH₄ = one medium dark circle joined to four small light circles · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
O₂ KEY
SiH₄
SiO₂
Neither reactant
STO-021 · hard · application · seq 18 · authored (excess-trap: 9.0 vs 2.0) · numeric
In a smelting furnace, 2PbO + C → 2Pb + CO₂. C is the limiting reactant with 2.0 mol available; the vessel also holds 9.0 mol of PbO. How many moles of Pb does the reaction produce? Give your answer in moles to 2 significant figures.
Answer: 4.0 mol KEY
also accepted: 4.0 mol
wrongValueDiagnostics
[
{
"value": 9.0,
"misconception": "you started from the excess reactant PbO — the limiting reactant C caps the product",
"feedback": null
},
{
"value": 1.0,
"misconception": "you flipped the mole ratio — the coefficient of Pb goes on top and C's underneath",
"feedback": null
},
{
"value": 2.0,
"misconception": "you carried the limiting reactant's moles straight across without the mole ratio",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of C (limiting) = 2.0 mol\nThe 9.0 mol of PbO is the excess reactant — it does not enter the calculation.\nWrite down the equation:\nmoles of Pb = moles of C × (coefficient of Pb ÷ coefficient of C)\nSubstitute in the values, and calculate:\nmoles of Pb = 2.0 × (2 ÷ 1)\nmoles of Pb = 4.0 mol"
}
STO-022 · hard · application · seq 19 · authored (mass chain; hydrogen reduction of copper oxide — classroom classic; smaller-mass reactant is NOT limiting, STO-019 echo) · numeric
31.8 g of CuO and 1.2 g of H₂ react by CuO + H₂ → Cu + H₂O. Molar masses (exact): CuO 79.5 g/mol, H₂ 2.0 g/mol, Cu 63.5 g/mol. Give the mass of Cu that forms, in grams (3 significant figures).
Answer: 25.4 g KEY
also accepted: 25.4 g
wrongValueDiagnostics
[
{
"value": 38.1,
"misconception": "you calculated from the excess reactant H₂ — using up 0.400 mol of CuO needs only 0.400 mol of H₂, and 0.60 mol is available, so CuO is limiting",
"feedback": null
},
{
"value": 33.0,
"misconception": "you added the two reactant masses — the leftover excess H₂ is not part of the asked product",
"feedback": null
},
{
"value": 31.8,
"misconception": "you used CuO's molar mass in the final step — the wanted substance Cu's molar mass is 63.5 g/mol",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of CuO = 31.8 g, M of CuO = 79.5 g/mol, m of H₂ = 1.2 g, M of H₂ = 2.0 g/mol, M of Cu = 63.5 g/mol\nStep 1 — convert each mass to moles:\nn(CuO) = 31.8 / 79.5 = 0.400 mol\nn(H₂) = 1.2 / 2.0 = 0.60 mol\nStep 2 — identify the limiting reactant:\nmoles of H₂ needed = 0.400 × (1 ÷ 1) = 0.400 mol — 0.60 mol is available, so CuO is limiting\nStep 3 — moles of product from the limiting reactant:\nmoles of Cu = 0.400 × (1 ÷ 1) = 0.400 mol\nStep 4 — convert the product moles to mass:\nm(Cu) = 0.400 × 63.5\nm(Cu) = 25.4 g"
}
STO-022 · hard · application · seq 20 · authored (plan surface: first-wrong-step spot on the two-candidate variant) · mcq4
For a masses-of-both-reactants problem, a student plans: (1) convert both masses to moles; (2) convert the moles of EACH reactant separately into moles of product; (3) convert the LARGER of the two product amounts to mass. Where is the first wrong step?
Step 3 — the limiting reactant permits only the SMALLER of the two amounts; the larger one assumes a supply that runs out. KEY
Step 1 — the masses should be compared directly without converting to moles.
Step 2 — moles of product can only ever be calculated from one fixed reactant, never from each in turn.
There is no wrong step — the plan is correct as written.
STO-020 · medium · application · seq 21 · authored (FRQ-1 form B: needed > available; strontium replaces chromium — Cr₂O₃ token collision with 18D) · frq
A sealed vessel holds 6.0 mol of Sr and 2.0 mol of O₂. The reaction is 2Sr + O₂ → 2SrO.
A sealed vessel holds 6.0 mol of Sr and 2.0 mol of O₂. The reaction is 2Sr + O₂ → 2SrO.
(a) Calculate how many moles of O₂ would be needed to react completely with all 6.0 mol of Sr. Show your working.
(b) Compare the moles of O₂ needed with the moles of O₂ available, and say which reactant is limiting.
(c) Name the excess reactant, and state what will be true of it when the reaction stops.
Part a. Calculate how many moles of O₂ would be needed to react completely with all 6.0 mol of Sr. Show your working.
Model: Amounts supplied: 6.0 mol of strontium, 2.0 mol of oxygen.
moles of oxygen needed = moles of strontium × (coefficient of oxygen ÷ coefficient of strontium)
moles of oxygen needed = 6.0 × (1 ÷ 2)
moles of oxygen needed = 3.0 mol
Part b. Compare the moles of O₂ needed with the moles of O₂ available, and say which reactant is limiting.
Model: 3.0 mol needed > 2.0 mol available — less oxygen is available than needed, so the oxygen runs out first. The oxygen is the limiting reactant.
Part c. Name the excess reactant, and state what will be true of it when the reaction stops.
Model: The excess reactant is the strontium. When the reaction stops, some of it will be left over, unreacted — it reacted only while the oxygen lasted.
Full model answer. (a) Amounts supplied: 6.0 mol of strontium, 2.0 mol of oxygen.
moles of oxygen needed = moles of strontium × (coefficient of oxygen ÷ coefficient of strontium)
moles of oxygen needed = 6.0 × (1 ÷ 2)
moles of oxygen needed = 3.0 mol
(b) 3.0 mol needed > 2.0 mol available — less oxygen is available than needed, so the oxygen runs out first. The oxygen is the limiting reactant.
(c) The excess reactant is the strontium. When the reaction stops, some of it will be left over, unreacted — it reacted only while the oxygen lasted.
STO-022 · hard · application · seq 22 · authored (FRQ-2 form B: smaller-mass reactant NOT limiting — STO-019 echo per WE2) · frq
A student heats 29.88 g of NiO in 1.4 g of H₂; the reaction is NiO + H₂ → Ni + H₂O.
A student heats 29.88 g of NiO in 1.4 g of H₂; the reaction is NiO + H₂ → Ni + H₂O.
(a) Calculate the moles of each reactant.
(b) Identify the limiting reactant, showing the comparison of needed moles against available moles.
(c) Calculate the mass of Ni that forms. Show every step of your working.
Molar masses (exact): NiO 74.7 g/mol, H₂ 2.0 g/mol, Ni 58.7 g/mol.
Part b. Identify the limiting reactant, showing the comparison of needed moles against available moles.
Model: moles of hydrogen needed = moles of nickel oxide × (coefficient ratio) = 0.40 mol
0.40 mol needed < 0.70 mol available — the hydrogen will not run out, so the nickel oxide is used up first: it is the limiting reactant.
Part c. Calculate the mass of Ni that forms. Show every step of your working.
Molar masses (exact): NiO 74.7 g/mol, H₂ 2.0 g/mol, Ni 58.7 g/mol.
Model: moles of nickel = 0.4000 × (1 ÷ 1) = 0.4000 mol (starting from the limiting reactant)
mass of nickel = 0.4000 × 58.7
mass of nickel = 23.48 g
Full model answer. (a) moles of nickel oxide = 29.88 / 74.7 = 0.4000 mol
moles of hydrogen = 1.4 / 2.0 = 0.70 mol
(b) moles of hydrogen needed = moles of nickel oxide × (coefficient ratio) = 0.40 mol
0.40 mol needed < 0.70 mol available — the hydrogen will not run out, so the nickel oxide is used up first: it is the limiting reactant.
(c) moles of nickel = 0.4000 × (1 ÷ 1) = 0.4000 mol (starting from the limiting reactant)
mass of nickel = 0.4000 × 58.7
mass of nickel = 23.48 g
One reactant of a reaction has been completely used up, but plenty of the other reactant remains. Can the remaining reactant keep making product by itself?
No — forming product needs every reactant, so the reaction has stopped. KEY
Yes — a reactant can always form product alone, just more slowly.
Yes, but only if the mixture is stirred so the leftover spreads out.
No — the leftover reactant is destroyed when the other one runs out.
In the needed-versus-available check for two reactants, MORE of the second reactant is available than would be needed to use up the first. Which reactant is limiting?
The first reactant — the one the check started from — runs out first. KEY
The second reactant — having more than needed means it runs out.
Neither reactant is limiting in this situation.
The roles cannot be assigned without the masses of the two reactants.
Chromium metal is heated with a measured amount of a second reactant until the reaction stops. Every trace of the chromium has been used up, and some of the second reactant remains. Which label fits the chromium?
It is the limiting reactant — it was completely used up first. KEY
It is the excess reactant — metals are usually supplied in excess.
It is the excess reactant, because the other reactant is what remains.
It cannot be labeled without knowing the balanced equation.
STO-014 · medium · application · seq 5 · authored (fixed-supply class; QC wave-3b rebuild: glow stick was rate-fading per sol loop 2; the candle replacement collided with STO-014's own lit-candle spec text (V-T7) — final scenario: baking-soda volcano, vinegar's acid = the fixed supply) · mcq4
A classroom volcano model erupts when vinegar is poured onto a mound of baking soda. The foaming soon stops, though a large mound of baking soda is still sitting in the model. Why did the foaming stop?
The vinegar's acid was completely used up, so the reaction stopped. KEY
The baking soda was completely used up.
The foam pressed down on the mound and blocked the powder from reacting.
Volcano models always foam for a fixed time, whatever is poured in.
The figure shows a reactant mixture for the reaction 2Ni + O₂ → 2NiO. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "four Ni atoms (each one medium gray circle, unjoined) and four O₂ molecules (each two large red circles joined), mixed together, widely spaced, no Ni touchi… · key: "Ni = one medium gray circle · O₂ = two large red circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four Ni atoms (each one medium gray circle, unjoined) and four O₂ molecules (each two large red circles joined), mixed together, widely spaced, no Ni touching an O₂",
"key": "Ni = one medium gray circle · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
The figure shows a reactant mixture for the reaction Si + 2Cl₂ → SiCl₄. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "three Si atoms (each one medium shaded circle, unjoined) and six Cl₂ molecules (each two large green circles joined), mixed together, widely spaced, no Si t… · key: "Si = one medium shaded circle · Cl₂ = two large green circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three Si atoms (each one medium shaded circle, unjoined) and six Cl₂ molecules (each two large green circles joined), mixed together, widely spaced, no Si touching a Cl₂",
"key": "Si = one medium shaded circle · Cl₂ = two large green circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
The figure shows a reactant mixture. It reacts by 2K + S → K₂S. Which particle diagram shows the mixture after the reaction stops?
FIGURE (stimulus · stimulus) — see raw specraw spec
{
"option_format": "image",
"render_contract": "R2 image-XOR-text: options render as pictures ONLY; the 'text'/'key' placeholder strings are internal ids for validation and are never rendered and never used for aria (R23 — descriptions must not exceed what a sighted student sees).",
"figureRequired": true,
"figureShape": "stimulus",
"stimulus": {
"figureType": "particle-diagram",
"box": "six K atoms (each one medium gray circle, unjoined) and two S atoms (each one small yellow circle, unjoined), mixed together, widely spaced, no K touching an S",
"key": "K = one medium gray circle · S = one small yellow circle",
"style": "clean textbook particle diagram, single labeled box"
},
"optionFigureSpecs": [
{
"option": "key",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two K₂S units (each two medium gray circles joined to one small yellow circle) and two K atoms (each one medium gray circle, unjoined), mixed together, widely spaced",
"key": "K₂S = two medium gray circles joined to one small yellow circle · K = one medium gray circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor1",
"figureSpec": {
"figureType": "particle-diagram",
"box": "one K₂S unit, one S atom (one small yellow circle), and four K atoms, mixed together, widely spaced",
"key": "K₂S = two medium gray circles joined to one small yellow circle · S = one small yellow circle · K = one medium gray circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor2",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three K₂S units (each two medium gray circles joined to one small yellow circle), no other particles, mixed together, widely spaced",
"key": "K₂S = two medium gray circles joined to one small yellow circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor3",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two K₂S units (each two medium gray circles joined to one small yellow circle) only, no other particles, mixed together, widely spaced",
"key": "K₂S = two medium gray circles joined to one small yellow circle",
"style": "clean textbook particle diagram, single labeled box"
}
}
]
}
Picture 1 KEY
Picture 2
Picture 3
Picture 4
STO-018 · medium · application · seq 9 · authored (ask class: why-the-limiting-amount-caps-product, MECH-LIMITING verbatim as the key; unnamed precipitation scenario) · mcq4
Two solutions are mixed and a solid product forms until one solution's dissolved reactant is completely used up. Why can no more solid form after that moment, even though the other reactant remains in the mixture?
Product can only form while every reactant is still available, so the reactant that runs out first stops product formation. KEY
Because the dissolved reactant is the most reactive substance in the mixture.
Because a solid product can only ever come from the reactant that was present in the smaller volume of solution.
Because the other reactant stops working once the solid starts to appear.
STO-019 · medium · application · seq 10 · authored (shortcut-fails family; boron combustion replaces 4Al+3O₂ — 'al and 5.0 mol' duplicates STO-020 WE2 tokens, and 'oxygen' is an own-lesson material) · mcq4
6.0 mol of B and 5.0 mol of O₂ react by 4B + 3O₂ → 2B₂O₃. Which reasoning correctly identifies the limiting reactant?
B — using all 6.0 mol of B needs only 4.5 mol of O₂, so the O₂ outlasts the B. KEY
O₂ — it is present in the smaller amount, and the smaller amount is always limiting.
O₂ — the reactant that feeds the flame always runs out first.
Neither — 6.0 mol against 5.0 mol matches the 4-to-3 coefficient ratio.
3.0 mol of CS₂ and 8.0 mol of O₂ react by CS₂ + 3O₂ → CO₂ + 2SO₂. Which reactant is the limiting reactant?
O₂ KEY
CS₂
SO₂
Neither reactant
STO-020 · medium · application · seq 12 · authored (exact-ratio case; tungsten chlorination replaces ethanol combustion — 'by c 2 h'/'2 react by c' token collisions with 19D/22D-E) · mcq4
2.0 mol of W and 6.0 mol of Cl₂ react by W + 3Cl₂ → WCl₆. Which statement about the limiting reactant is correct?
Neither — the two reactants run out together. KEY
W is the limiting reactant.
Cl₂ is the limiting reactant.
It cannot be decided from the moles alone — the masses of the two reactants are needed as well.
STO-019 · medium · transfer · seq 13 · authored (claim class: equal-moles-neither; instance-free) · mcq4
A margin note in a borrowed textbook reads: 'If you start with equal moles of the two reactants, neither one is limiting.' Which reply is correct?
Wrong — unless the coefficients are equal too, one reactant still runs out first. KEY
Right — equal supplies of the two reactants are always used up at the same moment.
Right, provided the two reactants have the same molar mass.
Wrong — with equal moles, BOTH reactants are limiting at once.
STO-018 · medium · transfer · seq 14 · authored (claim class: cannot-be-finished-while-reactant-remains) · mcq4
A student argues: 'A reaction cannot really be finished while any reactant remains — it must still be creeping along using the leftover.' Which reply is correct?
Wrong — with one reactant completely gone, product formation has fully ended; the leftover has nothing to react with. KEY
Right — every reaction keeps creeping along for as long as any starting material at all remains.
Right — the creeping is just too slow to notice.
Wrong — reactions finish only when the product decides to stop forming.
STO-017 · medium · application · seq 15 · authored (exact-ratio family; the seq-19 seed's freed reaction reused here as a fresh figure instance — gate items are not corpus) · mcq4
The figure shows a reactant mixture for the reaction C + O₂ → CO₂. Which particle diagram shows the mixture after the reaction stops?
FIGURE (stimulus · stimulus) — see raw specraw spec
{
"option_format": "image",
"render_contract": "R2 image-XOR-text: options render as pictures ONLY; the 'text'/'key' placeholder strings are internal ids for validation and are never rendered and never used for aria (R23 — descriptions must not exceed what a sighted student sees).",
"figureRequired": true,
"figureShape": "stimulus",
"stimulus": {
"figureType": "particle-diagram",
"box": "four C atoms (each one medium black circle, unjoined) and four O₂ molecules (each two large red circles joined), mixed together, widely spaced, no C touching an O₂",
"key": "C = one medium black circle · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
},
"optionFigureSpecs": [
{
"option": "key",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four CO₂ molecules (each one medium black circle joined to two large red circles), mixed together, widely spaced, no other particles",
"key": "CO₂ = one medium black circle joined to two large red circles",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor1",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three CO₂ molecules, one C atom (one medium black circle), and one O₂ molecule (two large red circles joined), mixed together, widely spaced",
"key": "CO₂ = one medium black circle joined to two large red circles · C = one medium black circle · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor2",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four product units drawn with the wrong recipe: each one medium black circle joined to one large red circle, plus four lone large red circles, mixed together, widely spaced",
"key": "product drawn as one medium black circle joined to one large red circle · lone large red circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor3",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four CO₂ molecules (each one medium black circle joined to two large red circles) and one O₂ molecule (two large red circles joined), mixed together, widely spaced",
"key": "CO₂ = one medium black circle joined to two large red circles · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
]
}
Picture 1 KEY
Picture 2
Picture 3
Picture 4
STO-021 · medium · application · seq 16 · authored (halving ratio; stem names aluminum/hydrochloric acid — 'aluminum' is a MATERIALS token absent from every scope corpus, word-laned to this seq) · numeric
Aluminum dissolves in hydrochloric acid: 2Al + 6HCl → 2AlCl₃ + 3H₂. HCl is the limiting reactant with 6.0 mol available, and 3.0 mol of Al is present. How many moles of H₂ form from this mixture? Give your answer in moles to 2 significant figures.
Answer: 3.0 mol KEY
also accepted: 3.0 mol
wrongValueDiagnostics
[
{
"value": 4.5,
"misconception": "you started from the excess reactant Al — the limiting reactant HCl caps the product",
"feedback": null
},
{
"value": 12.0,
"misconception": "you flipped the mole ratio — the coefficient of H₂ goes on top and HCl's underneath",
"feedback": null
},
{
"value": 6.0,
"misconception": "you carried the limiting reactant's moles straight across without the mole ratio",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of HCl (limiting) = 6.0 mol\nThe 3.0 mol of Al is the excess reactant — it does not enter the calculation.\nWrite down the equation:\nmoles of H₂ = moles of HCl × (coefficient of H₂ ÷ coefficient of HCl)\nSubstitute in the values, and calculate:\nmoles of H₂ = 6.0 × (3 ÷ 6)\nmoles of H₂ = 3.0 mol"
}
STO-016 · medium · transfer · seq 17 · authored (bigger-pile-limiting family, fresh reaction) · mcq4
The figure shows a reactant mixture for the reaction C₂H₄ + 3O₂ → 2CO₂ + 2H₂O. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "three C₂H₄ molecules (each two medium black circles joined side by side, each black circle also joined to two small light circles) and six O₂ molecules (eac… · key: "C₂H₄ = two medium black circles joined, each also joined to two small light circles · O₂ = two large red circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three C₂H₄ molecules (each two medium black circles joined side by side, each black circle also joined to two small light circles) and six O₂ molecules (each two large red circles joined), mixed together, widely spaced, no two different molecules touching",
"key": "C₂H₄ = two medium black circles joined, each also joined to two small light circles · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
O₂ KEY
C₂H₄
CO₂
Neither reactant
STO-021 · hard · application · seq 18 · authored (excess-trap with a one-to-one wanted ratio; Kroll-process instance) · numeric
In the Kroll process, TiCl₄ + 2Mg → Ti + 2MgCl₂. TiCl₄ is the limiting reactant with 2.0 mol available; the vessel also holds 9.0 mol of Mg. How many moles of Ti does the reaction produce? Give your answer in moles to 2 significant figures.
Answer: 2.0 mol KEY
also accepted: 2.0 mol
wrongValueDiagnostics
[
{
"value": 4.5,
"misconception": "you started from the excess reactant Mg — the limiting reactant TiCl₄ caps the product",
"feedback": null
},
{
"value": 9.0,
"misconception": "you carried the excess reactant's moles across — the calculation must start from the limiting reactant TiCl₄",
"feedback": null
},
{
"value": 4.0,
"misconception": "you used MgCl₂'s coefficient — the wanted substance is Ti, whose coefficient is 1",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of TiCl₄ (limiting) = 2.0 mol\nThe 9.0 mol of Mg is the excess reactant — it does not enter the calculation.\nWrite down the equation:\nmoles of Ti = moles of TiCl₄ × (coefficient of Ti ÷ coefficient of TiCl₄)\nSubstitute in the values, and calculate:\nmoles of Ti = 2.0 × (1 ÷ 1)\nmoles of Ti = 2.0 mol"
}
STO-022 · hard · application · seq 19 · authored (mass chain; QC wave-3 rebuild: H₂+I₂⇌2HI is a classic equilibrium system per sol loop 2 — replaced with complete 2K + Cl₂ → 2KCl; ×2 product ratio kept) · numeric
19.55 g of K and 14.18 g of Cl₂ react on heating by 2K + Cl₂ → 2KCl. Molar masses (exact): K 39.1 g/mol, Cl₂ 70.9 g/mol, KCl 74.6 g/mol. Give the mass of KCl that forms, in grams (4 significant figures).
Answer: 29.84 g KEY
also accepted: 29.84 g
wrongValueDiagnostics
[
{
"value": 37.3,
"misconception": "you calculated from the excess reactant K — using up 0.5000 mol of K needs 0.2500 mol of Cl₂, and only 0.2000 mol is available, so Cl₂ is limiting",
"feedback": null
},
{
"value": 14.92,
"misconception": "you skipped the mole-ratio step — each mole of Cl₂ makes TWO moles of KCl",
"feedback": null
},
{
"value": 33.73,
"misconception": "you added the two reactant masses — the leftover excess K is not part of the product",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of K = 19.55 g, M of K = 39.1 g/mol, m of Cl₂ = 14.18 g, M of Cl₂ = 70.9 g/mol, M of KCl = 74.6 g/mol\nStep 1 — convert each mass to moles:\nn(K) = 19.55 / 39.1 = 0.5000 mol\nn(Cl₂) = 14.18 / 70.9 = 0.2000 mol\nStep 2 — identify the limiting reactant:\nmoles of Cl₂ needed = 0.5000 × (1 ÷ 2) = 0.2500 mol — only 0.2000 mol is available, so Cl₂ is limiting\nStep 3 — moles of product from the limiting reactant:\nmoles of KCl = 0.2000 × (2 ÷ 1) = 0.4000 mol\nStep 4 — convert the product moles to mass:\nm(KCl) = 0.4000 × 74.6\nm(KCl) = 29.84 g"
}
In the masses-given product calculation, why can the limiting-reactant comparison not be skipped, even when both reactant masses are known exactly?
Because without it the calculation may draw on reactant that runs out — the scarcer supply caps the product. KEY
It can be skipped — knowing both masses exactly is the same as knowing the limiting reactant.
Because the comparison is what converts grams into moles.
Because without it the answer comes out with the wrong unit.
STO-020 · medium · application · seq 21 · authored (FRQ-1 form C: needed < available branch — the picked-first reactant is limiting) · frq
A sealed vessel holds 6.0 mol of Ca and 4.0 mol of O₂. The reaction is 2Ca + O₂ → 2CaO.
A sealed vessel holds 6.0 mol of Ca and 4.0 mol of O₂. The reaction is 2Ca + O₂ → 2CaO.
(a) Calculate how many moles of O₂ would be needed to react completely with all 6.0 mol of Ca. Show your working.
(b) Compare the moles of O₂ needed with the moles of O₂ available, and say which reactant is limiting.
(c) Name the excess reactant, and state what will be true of it when the reaction stops.
Part a. Calculate how many moles of O₂ would be needed to react completely with all 6.0 mol of Ca. Show your working.
Model: Amounts supplied: 6.0 mol of calcium, 4.0 mol of oxygen.
moles of oxygen needed = moles of calcium × (coefficient of oxygen ÷ coefficient of calcium)
moles of oxygen needed = 6.0 × (1 ÷ 2)
moles of oxygen needed = 3.0 mol
Part b. Compare the moles of O₂ needed with the moles of O₂ available, and say which reactant is limiting.
Model: 3.0 mol needed < 4.0 mol available — the oxygen will not run out, so the calcium is used up first. The calcium is the limiting reactant.
Part c. Name the excess reactant, and state what will be true of it when the reaction stops.
Model: The excess reactant is the oxygen. When the reaction stops, some of it will be left over, unreacted — it reacted only while the calcium lasted.
Full model answer. (a) Amounts supplied: 6.0 mol of calcium, 4.0 mol of oxygen.
moles of oxygen needed = moles of calcium × (coefficient of oxygen ÷ coefficient of calcium)
moles of oxygen needed = 6.0 × (1 ÷ 2)
moles of oxygen needed = 3.0 mol
(b) 3.0 mol needed < 4.0 mol available — the oxygen will not run out, so the calcium is used up first. The calcium is the limiting reactant.
(c) The excess reactant is the oxygen. When the reaction stops, some of it will be left over, unreacted — it reacted only while the calcium lasted.
STO-022 · hard · application · seq 22 · authored (FRQ-2 form C: precipitation; larger-mole reactant IS limiting) · frq
A student stirs together two solutions, one holding 66.24 g of Pb(NO₃)₂ and the other 49.80 g of KI; the reaction is Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃.
A student stirs together two solutions, one holding 66.24 g of Pb(NO₃)₂ and the other 49.80 g of KI; the reaction is Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃.
(a) Calculate the moles of each reactant.
(b) Identify the limiting reactant, showing the comparison of needed moles against available moles.
(c) Calculate the mass of PbI₂ that forms. Show every step of your working.
Molar masses (exact): Pb(NO₃)₂ 331.2 g/mol, KI 166.0 g/mol, PbI₂ 461.0 g/mol.
Part a. Calculate the moles of each reactant.
Model: moles of lead nitrate = 66.24 / 331.2 = 0.2000 mol
moles of potassium iodide = 49.80 / 166.0 = 0.3000 mol
Part b. Identify the limiting reactant, showing the comparison of needed moles against available moles.
Model: moles of potassium iodide needed = moles of lead nitrate × (coefficient ratio) = 0.40 mol
0.40 mol needed > 0.30 mol available — the potassium iodide falls short and runs out first: it is the limiting reactant.
Part c. Calculate the mass of PbI₂ that forms. Show every step of your working.
Molar masses (exact): Pb(NO₃)₂ 331.2 g/mol, KI 166.0 g/mol, PbI₂ 461.0 g/mol.
Model: moles of lead iodide = 0.3000 × (1 ÷ 2) = 0.1500 mol (starting from the limiting reactant)
mass of lead iodide = 0.1500 × 461.0
mass of lead iodide = 69.15 g
Full model answer. (a) moles of lead nitrate = 66.24 / 331.2 = 0.2000 mol
moles of potassium iodide = 49.80 / 166.0 = 0.3000 mol
(b) moles of potassium iodide needed = moles of lead nitrate × (coefficient ratio) = 0.40 mol
0.40 mol needed > 0.30 mol available — the potassium iodide falls short and runs out first: it is the limiting reactant.
(c) moles of lead iodide = 0.3000 × (1 ÷ 2) = 0.1500 mol (starting from the limiting reactant)
mass of lead iodide = 0.1500 × 461.0
mass of lead iodide = 69.15 g
Why do the coefficients of the balanced equation have to enter the limiting-reactant decision?
They set how many moles of one reactant are spent per mole of the other — the scaling the comparison needs. KEY
They do not have to enter — comparing the two mole amounts directly gives the same answer.
They give the number of grams of each reactant that react.
They show which reactant is more reactive, and the more reactive one is limiting.
STO-015 · easy · application · seq 4 · authored (evidence class: both-labels assign, pool[3] surface class with a fresh instance; word lane: calcium) · mcq4
Calcium metal reacts in a flask until the reaction stops on its own. Afterward, no calcium remains, but the liquid it reacted with is still present. Which statement assigns the roles correctly?
The calcium is the limiting reactant and the liquid is the excess reactant. KEY
The liquid is the limiting reactant and the calcium is the excess reactant.
Both the calcium and the liquid are limiting reactants.
The calcium is the limiting reactant, and the liquid holds no role because some remains.
STO-014 · medium · application · seq 5 · authored (fixed-supply class: small measured liquid; 'acid'/'tablet' are non-material words) · mcq4
An effervescent tablet is dropped into a small cup of acid and fizzes. The fizzing stops while most of the tablet is still there. Why did the fizzing stop?
The acid in the cup was completely used up. KEY
The tablet was completely used up.
The fizzing gas pushed the acid away from the tablet.
The figure shows a reactant mixture for the reaction 2Ag + S → Ag₂S. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "six Ag atoms (each one medium gray circle, unjoined) and six S atoms (each one small yellow circle, unjoined), mixed together, widely spaced, no Ag touching… · key: "Ag = one medium gray circle · S = one small yellow circle" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "six Ag atoms (each one medium gray circle, unjoined) and six S atoms (each one small yellow circle, unjoined), mixed together, widely spaced, no Ag touching an S",
"key": "Ag = one medium gray circle · S = one small yellow circle",
"style": "clean textbook particle diagram, single labeled box"
}
}
The figure shows a reactant mixture for the reaction 2Na + Br₂ → 2NaBr. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "four Na atoms (each one medium gray circle, unjoined) and two Br₂ molecules (each two large brown circles joined), mixed together, widely spaced, no Na touc… · key: "Na = one medium gray circle · Br₂ = two large brown circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four Na atoms (each one medium gray circle, unjoined) and two Br₂ molecules (each two large brown circles joined), mixed together, widely spaced, no Na touching a Br₂",
"key": "Na = one medium gray circle · Br₂ = two large brown circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
Neither reactant KEY
Na
Br₂
NaBr
STO-017 · medium · application · seq 8 · authored (leftover-excess family; Ni + Br₂ replaces Si + O₂ — 'o 2 sio 2' and 'for the reaction si' token collisions with 17B/7C) · mcq4
The figure shows a reactant mixture. It reacts by Ni + Br₂ → NiBr₂. Which particle diagram shows the mixture after the reaction stops?
FIGURE (stimulus · stimulus) — see raw specraw spec
{
"option_format": "image",
"render_contract": "R2 image-XOR-text: options render as pictures ONLY; the 'text'/'key' placeholder strings are internal ids for validation and are never rendered and never used for aria (R23 — descriptions must not exceed what a sighted student sees).",
"figureRequired": true,
"figureShape": "stimulus",
"stimulus": {
"figureType": "particle-diagram",
"box": "three Ni atoms (each one medium gray circle, unjoined) and five Br₂ molecules (each two large brown circles joined), mixed together, widely spaced, no Ni touching a Br₂",
"key": "Ni = one medium gray circle · Br₂ = two large brown circles joined",
"style": "clean textbook particle diagram, single labeled box"
},
"optionFigureSpecs": [
{
"option": "key",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three NiBr₂ units (each one medium gray circle joined to two large brown circles) and two Br₂ molecules (each two large brown circles joined), mixed together, widely spaced",
"key": "NiBr₂ = one medium gray circle joined to two large brown circles · Br₂ = two large brown circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor1",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two NiBr₂ units, one lone Ni atom (one medium gray circle), and three Br₂ molecules, mixed together, widely spaced",
"key": "NiBr₂ = one medium gray circle joined to two large brown circles · Ni = one medium gray circle · Br₂ = two large brown circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor2",
"figureSpec": {
"figureType": "particle-diagram",
"box": "five NiBr₂ units (each one medium gray circle joined to two large brown circles), no other particles, mixed together, widely spaced",
"key": "NiBr₂ = one medium gray circle joined to two large brown circles",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor3",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three NiBr₂ units (each one medium gray circle joined to two large brown circles) only, no other particles, mixed together, widely spaced",
"key": "NiBr₂ = one medium gray circle joined to two large brown circles",
"style": "clean textbook particle diagram, single labeled box"
}
}
]
}
Picture 1 KEY
Picture 2
Picture 3
Picture 4
STO-018 · medium · application · seq 9 · authored (ask class: which-change-makes-more-product; word lanes lead/bromine) · mcq4
A coil of lead reacts with bromine vapor in a sealed flask until the bromine is used up; some lead remains. Which change would let the reaction make MORE product?
Adding more bromine to the flask. KEY
Adding more lead to the flask.
Warming the flask so the leftover lead reacts by itself.
4.0 mol of CH₃OH and 7.0 mol of O₂ react by 2CH₃OH + 3O₂ → 2CO₂ + 4H₂O. Which reactant is the limiting reactant?
CH₃OH KEY
O₂
CO₂
Neither reactant
STO-020 · medium · application · seq 12 · authored (decimals case; tin-oxide smelting; ledgered decimals 0.30/0.40 printed at this seq only) · mcq4
0.30 mol of SnO₂ and 0.40 mol of C react by SnO₂ + 2C → Sn + 2CO. Which reactant is the limiting reactant?
C KEY
SnO₂
Sn
Neither reactant
STO-019 · medium · transfer · seq 13 · authored (claim class: amounts-alone; instance-free) · mcq4
A student claims: 'You can spot the limiting reactant from the two amounts alone — you never need the balanced equation.' Which reply is correct?
Wrong — the equation's coefficients set how fast each supply is spent, so without them the amounts cannot be compared fairly. KEY
Right — whichever amount is smaller runs out first, equation or no equation.
Right, as long as the amounts are given in moles rather than grams.
Wrong — the amounts are completely irrelevant; the balanced equation by itself names the limiting reactant.
STO-018 · medium · transfer · seq 14 · authored (claim class: stir-or-heat-restarts-it) · mcq4
A lab note reads: 'Our reaction stopped early. Stirring and gently heating the mixture should squeeze out more product, since one reactant is still left over.' Which reply is correct?
Wrong — if the other reactant is used up, stirring and warmth cannot make product from the leftover alone. KEY
Right — stirring brings leftover particles together so they can react with each other.
Right — heat can substitute for a used-up reactant.
Wrong — stirring actually destroys the product that has already formed.
The figure shows a reactant mixture for the reaction 2Li + Cl₂ → 2LiCl. Which particle diagram shows the mixture after the reaction stops?
FIGURE (stimulus · stimulus) — see raw specraw spec
{
"option_format": "image",
"render_contract": "R2 image-XOR-text: options render as pictures ONLY; the 'text'/'key' placeholder strings are internal ids for validation and are never rendered and never used for aria (R23 — descriptions must not exceed what a sighted student sees).",
"figureRequired": true,
"figureShape": "stimulus",
"stimulus": {
"figureType": "particle-diagram",
"box": "four Li atoms (each one small gray circle, unjoined) and two Cl₂ molecules (each two large green circles joined), mixed together, widely spaced, no Li touching a Cl₂",
"key": "Li = one small gray circle · Cl₂ = two large green circles joined",
"style": "clean textbook particle diagram, single labeled box"
},
"optionFigureSpecs": [
{
"option": "key",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four LiCl units (each one small gray circle joined to one large green circle), mixed together, widely spaced, no other particles",
"key": "LiCl = one small gray circle joined to one large green circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor1",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two LiCl units (each one small gray circle joined to one large green circle), no other particles, mixed together, widely spaced",
"key": "LiCl = one small gray circle joined to one large green circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor2",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two LiCl units, two Li atoms (each one small gray circle), and one Cl₂ molecule (two large green circles joined), mixed together, widely spaced",
"key": "LiCl = one small gray circle joined to one large green circle · Li = one small gray circle · Cl₂ = two large green circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor3",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two product units drawn with the wrong recipe: each one small gray circle joined to two large green circles, no other particles, mixed together, widely spaced",
"key": "product drawn as one small gray circle joined to two large green circles",
"style": "clean textbook particle diagram, single labeled box"
}
}
]
}
Picture 1 KEY
Picture 2
Picture 3
Picture 4
STO-021 · medium · application · seq 16 · authored (ratio-down case; roasting instance; 'copper' word removed — cross-scope material warn) · numeric
In a smelting works, Cu₂S ore is roasted by 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂. O₂ is the limiting reactant with 6.0 mol available, and 5.0 mol of Cu₂S is present. How many moles of Cu₂O form from this mixture? Give your answer in moles to 2 significant figures.
Answer: 4.0 mol KEY
also accepted: 4.0 mol
wrongValueDiagnostics
[
{
"value": 5.0,
"misconception": "you started from the excess reactant Cu₂S — the limiting reactant O₂ caps the product",
"feedback": null
},
{
"value": 9.0,
"misconception": "you flipped the mole ratio — the coefficient of Cu₂O goes on top and O₂'s underneath",
"feedback": null
},
{
"value": 6.0,
"misconception": "you carried the limiting reactant's moles straight across without the mole ratio",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of O₂ (limiting) = 6.0 mol\nThe 5.0 mol of Cu₂S is the excess reactant — it does not enter the calculation.\nWrite down the equation:\nmoles of Cu₂O = moles of O₂ × (coefficient of Cu₂O ÷ coefficient of O₂)\nSubstitute in the values, and calculate:\nmoles of Cu₂O = 6.0 × (2 ÷ 3)\nmoles of Cu₂O = 4.0 mol"
}
STO-016 · medium · transfer · seq 17 · authored (bigger-pile-limiting family, 4:1 demand) · mcq4
The figure shows a reactant mixture for the reaction 4Li + O₂ → 2Li₂O. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "eight Li atoms (each one small gray circle, unjoined) and four O₂ molecules (each two large red circles joined), mixed together, widely spaced, no Li touchi… · key: "Li = one small gray circle · O₂ = two large red circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "eight Li atoms (each one small gray circle, unjoined) and four O₂ molecules (each two large red circles joined), mixed together, widely spaced, no Li touching an O₂",
"key": "Li = one small gray circle · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
Li KEY
O₂
Li₂O
Neither reactant
STO-021 · hard · application · seq 18 · authored (excess-trap: 9.0 vs 2.0, doubling ratio) · numeric
In a thermite-type reaction, Cr₂O₃ + 2Al → 2Cr + Al₂O₃. Cr₂O₃ is the limiting reactant with 2.0 mol available; the vessel also holds 9.0 mol of Al. How many moles of Cr does the reaction produce? Give your answer in moles to 2 significant figures.
Answer: 4.0 mol KEY
also accepted: 4.0 mol
wrongValueDiagnostics
[
{
"value": 9.0,
"misconception": "you started from the excess reactant Al — the limiting reactant Cr₂O₃ caps the product",
"feedback": null
},
{
"value": 1.0,
"misconception": "you flipped the mole ratio — the coefficient of Cr goes on top and Cr₂O₃'s underneath",
"feedback": null
},
{
"value": 2.0,
"misconception": "you carried the limiting reactant's moles straight across without the mole ratio",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of Cr₂O₃ (limiting) = 2.0 mol\nThe 9.0 mol of Al is the excess reactant — it does not enter the calculation.\nWrite down the equation:\nmoles of Cr = moles of Cr₂O₃ × (coefficient of Cr ÷ coefficient of Cr₂O₃)\nSubstitute in the values, and calculate:\nmoles of Cr = 2.0 × (2 ÷ 1)\nmoles of Cr = 4.0 mol"
}
STO-022 · hard · application · seq 19 · authored (mass chain; silver-oxide reduction replaces ethene hydrogenation — 'c 2 h 4'/'2 c 2 h' token collisions with 17C/11A) · numeric
46.34 g of Ag₂O and 1.0 g of H₂ react on heating by Ag₂O + H₂ → 2Ag + H₂O. Molar masses (exact): Ag₂O 231.7 g/mol, H₂ 2.0 g/mol, Ag 107.9 g/mol. Give the mass of Ag that forms, in grams (4 significant figures).
Answer: 43.16 g KEY
also accepted: 43.16 g
wrongValueDiagnostics
[
{
"value": 107.9,
"misconception": "you calculated from the excess reactant H₂ — using up 0.20 mol of Ag₂O needs only 0.20 mol of H₂, and 0.50 mol is available, so Ag₂O is limiting",
"feedback": null
},
{
"value": 21.58,
"misconception": "you skipped the mole-ratio step — each mole of Ag₂O releases TWO moles of Ag",
"feedback": null
},
{
"value": 47.34,
"misconception": "you added the two reactant masses — the leftover excess H₂ is not part of the product",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Ag₂O = 46.34 g, M of Ag₂O = 231.7 g/mol, m of H₂ = 1.0 g, M of H₂ = 2.0 g/mol, M of Ag = 107.9 g/mol\nStep 1 — convert each mass to moles:\nn(Ag₂O) = 46.34 / 231.7 = 0.2000 mol\nn(H₂) = 1.0 / 2.0 = 0.50 mol\nStep 2 — identify the limiting reactant:\nmoles of H₂ needed = 0.2000 × (1 ÷ 1) = 0.2000 mol — 0.50 mol is available, so Ag₂O is limiting\nStep 3 — moles of product from the limiting reactant:\nmoles of Ag = 0.2000 × (2 ÷ 1) = 0.4000 mol\nStep 4 — convert the product moles to mass:\nm(Ag) = 0.4000 × 107.9\nm(Ag) = 43.16 g"
}
Arrange the masses-given product calculation correctly. Which ordering of the four links is right?
Masses to moles for both reactants → limiting-reactant comparison → mole ratio from the limiting reactant to the product → product moles to mass. KEY
Limiting-reactant comparison → masses to moles for both reactants → mole ratio to the product → product moles to mass.
Masses to moles for both reactants → mole ratio from each reactant separately → limiting-reactant comparison → add the two mass answers.
Masses to moles → limiting-reactant comparison → product moles to mass → mole ratio.
STO-020 · medium · application · seq 21 · authored (FRQ-1 form D: hexane, 1.0/6.0 — '2.0 mol of c' gram collision with 11A resolved; 19/14 print at this seq only) · frq
A sealed vessel holds 1.0 mol of C₆H₁₄ and 6.0 mol of O₂. The reaction is 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O.
A sealed vessel holds 1.0 mol of C₆H₁₄ and 6.0 mol of O₂. The reaction is 2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O.
(a) Calculate how many moles of O₂ would be needed to react completely with all 1.0 mol of C₆H₁₄. Show your working.
(b) Compare the moles of O₂ needed with the moles of O₂ available, and say which reactant is limiting.
(c) Name the excess reactant, and state what will be true of it when the reaction stops.
Part a. Calculate how many moles of O₂ would be needed to react completely with all 1.0 mol of C₆H₁₄. Show your working.
Model: Amounts supplied: 1.0 mol of hexane, 6.0 mol of oxygen.
moles of oxygen needed = moles of hexane × (coefficient of oxygen ÷ coefficient of hexane)
moles of oxygen needed = 1.0 × (19 ÷ 2)
moles of oxygen needed = 9.5 mol
Part b. Compare the moles of O₂ needed with the moles of O₂ available, and say which reactant is limiting.
Model: 9.5 mol needed > 6.0 mol available — less oxygen is available than needed, so the oxygen runs out first. The oxygen is the limiting reactant.
Part c. Name the excess reactant, and state what will be true of it when the reaction stops.
Model: The excess reactant is the hexane. When the reaction stops, some of it will be left over, unreacted — it reacted only while the oxygen lasted.
Full model answer. (a) Amounts supplied: 1.0 mol of hexane, 6.0 mol of oxygen.
moles of oxygen needed = moles of hexane × (coefficient of oxygen ÷ coefficient of hexane)
moles of oxygen needed = 1.0 × (19 ÷ 2)
moles of oxygen needed = 9.5 mol
(b) 9.5 mol needed > 6.0 mol available — less oxygen is available than needed, so the oxygen runs out first. The oxygen is the limiting reactant.
(c) The excess reactant is the hexane. When the reaction stops, some of it will be left over, unreacted — it reacted only while the oxygen lasted.
STO-022 · hard · application · seq 22 · authored (FRQ-2 form D: pentene hydrogenation — propene's C₃H₈ product token collided with the seq-16A propane lane; catalyst framing kept) · frq
A chemist passes 21.03 g of C₅H₁₀ and 1.6 g of H₂ over a catalyst; the reaction is C₅H₁₀ + H₂ → C₅H₁₂.
A chemist passes 21.03 g of C₅H₁₀ and 1.6 g of H₂ over a catalyst; the reaction is C₅H₁₀ + H₂ → C₅H₁₂.
(a) Calculate the moles of each reactant.
(b) Identify the limiting reactant, showing the comparison of needed moles against available moles.
(c) Calculate the mass of C₅H₁₂ that forms. Show every step of your working.
Molar masses (exact): C₅H₁₀ 70.1 g/mol, H₂ 2.0 g/mol, C₅H₁₂ 72.2 g/mol.
Part b. Identify the limiting reactant, showing the comparison of needed moles against available moles.
Model: moles of hydrogen needed = moles of pentene × (coefficient ratio) = 0.30 mol
0.30 mol needed < 0.80 mol available — the hydrogen will not run out, so the pentene is used up first: it is the limiting reactant.
Part c. Calculate the mass of C₅H₁₂ that forms. Show every step of your working.
Molar masses (exact): C₅H₁₀ 70.1 g/mol, H₂ 2.0 g/mol, C₅H₁₂ 72.2 g/mol.
Model: moles of pentane = 0.3000 × (1 ÷ 1) = 0.3000 mol (starting from the limiting reactant)
mass of pentane = 0.3000 × 72.2
mass of pentane = 21.66 g
Full model answer. (a) moles of pentene = 21.03 / 70.1 = 0.3000 mol
moles of hydrogen = 1.6 / 2.0 = 0.80 mol
(b) moles of hydrogen needed = moles of pentene × (coefficient ratio) = 0.30 mol
0.30 mol needed < 0.80 mol available — the hydrogen will not run out, so the pentene is used up first: it is the limiting reactant.
(c) moles of pentane = 0.3000 × (1 ÷ 1) = 0.3000 mol (starting from the limiting reactant)
mass of pentane = 0.3000 × 72.2
mass of pentane = 21.66 g
When a reaction stops, some of the excess reactant remains in the mixture. What did that remaining portion do during the reaction?
Nothing — it is the surplus that never found a partner to react with. KEY
It reacted and then turned back.
All of the excess reactant stayed completely untouched from the very start.
It became part of the product but can fall back out.
STO-020 · easy · recall · seq 3 · authored (recall probe E: which-equation-does-the-work; displayEquation verbatim is the taught rule line — canonical wording, whitelist-eligible; flagged for the fix loop if V-T7 disagrees) · mcq4
Which calculation does the work in the needed-versus-available check?
The mole-ratio equation — the same one used for every mole-to-mole conversion. KEY
The mass equation m = n × M applied to each reactant.
Dividing the larger of the two mole amounts by the smaller of the two.
Adding the two mole amounts together.
STO-015 · easy · application · seq 4 · authored (evidence class: limiting pick from stopped-plus-leftover, pool[5] surface class with a fresh instance; word lane: cobalt) · mcq4
Cobalt powder is stirred into a solution and reacts until the fizzing stops. The dissolved compound in the solution has been used up completely; cobalt powder remains. Which substance is the limiting reactant?
The dissolved reactant in the solution. KEY
The cobalt powder.
The product that formed during the fizzing.
Neither substance, because the reaction stopped on its own.
STO-014 · medium · application · seq 5 · authored (fixed-supply class: sealed pouch — spare pool[2]'s scenario class with the named substances removed; 'metal'/'gas' generic) · mcq4
A sealed warming pouch heats up as the metal powder inside reacts with a small amount of gas sealed into the pouch. The reaction stops and the pouch goes cold while plenty of metal powder remains. Why did the reaction stop?
The pouch's sealed gas supply was completely used up. KEY
The metal powder was completely used up.
The heat leaked away faster than the reaction could make it.
The metal became too cold to react.
STO-016 · medium · application · seq 6 · authored (equal-counts family, demand direction flipped — the OTHER reactant loses) · mcq4
The figure shows a reactant mixture for the reaction 2Al + 3I₂ → 2AlI₃. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "six Al atoms (each one medium gray circle, unjoined) and six I₂ molecules (each two large purple circles joined), mixed together, widely spaced, no Al touch… · key: "Al = one medium gray circle · I₂ = two large purple circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "six Al atoms (each one medium gray circle, unjoined) and six I₂ molecules (each two large purple circles joined), mixed together, widely spaced, no Al touching an I₂",
"key": "Al = one medium gray circle · I₂ = two large purple circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
The figure shows a reactant mixture for the reaction N₂ + 2O₂ → 2NO₂. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "two N₂ molecules (each two medium blue circles joined) and four O₂ molecules (each two large red circles joined), mixed together, widely spaced, no two diff… · key: "N₂ = two medium blue circles joined · O₂ = two large red circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two N₂ molecules (each two medium blue circles joined) and four O₂ molecules (each two large red circles joined), mixed together, widely spaced, no two different molecules touching",
"key": "N₂ = two medium blue circles joined · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
The figure shows a reactant mixture that reacts as 2Al + 3S → Al₂S₃. Which particle diagram shows the mixture after the reaction stops?
FIGURE (stimulus · stimulus) — see raw specraw spec
{
"option_format": "image",
"render_contract": "R2 image-XOR-text: options render as pictures ONLY; the 'text'/'key' placeholder strings are internal ids for validation and are never rendered and never used for aria (R23 — descriptions must not exceed what a sighted student sees).",
"figureRequired": true,
"figureShape": "stimulus",
"stimulus": {
"figureType": "particle-diagram",
"box": "four Al atoms (each one medium gray circle, unjoined) and eight S atoms (each one small yellow circle, unjoined), mixed together, widely spaced, no Al touching an S",
"key": "Al = one medium gray circle · S = one small yellow circle",
"style": "clean textbook particle diagram, single labeled box"
},
"optionFigureSpecs": [
{
"option": "key",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two Al₂S₃ units (each two medium gray circles joined to three small yellow circles) and two S atoms (each one small yellow circle, unjoined), mixed together, widely spaced",
"key": "Al₂S₃ = two medium gray circles joined to three small yellow circles · S = one small yellow circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor1",
"figureSpec": {
"figureType": "particle-diagram",
"box": "one Al₂S₃ unit, two Al atoms (each one medium gray circle), and five S atoms, mixed together, widely spaced",
"key": "Al₂S₃ = two medium gray circles joined to three small yellow circles · Al = one medium gray circle · S = one small yellow circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor2",
"figureSpec": {
"figureType": "particle-diagram",
"box": "three Al₂S₃ units (each two medium gray circles joined to three small yellow circles), no other particles, mixed together, widely spaced",
"key": "Al₂S₃ = two medium gray circles joined to three small yellow circles",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor3",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two Al₂S₃ units (each two medium gray circles joined to three small yellow circles) only, no other particles, mixed together, widely spaced",
"key": "Al₂S₃ = two medium gray circles joined to three small yellow circles",
"style": "clean textbook particle diagram, single labeled box"
}
}
]
}
Picture 1 KEY
Picture 2
Picture 3
Picture 4
STO-018 · medium · application · seq 9 · authored (ask class: add-more-excess with a visible-evidence twist; word lane: chalk; 'acid' non-material) · mcq4
Chalk chips fizz in a cup of acid until the fizzing stops; unreacted chips remain. A student adds MORE CHALK CHIPS, expecting more fizzing. What actually happens, and why?
No new fizzing — the acid is used up, and the added chips have nothing to react with. KEY
Vigorous fizzing returns, because fresh solid surface restarts any reaction.
Gentle fizzing returns, using the acid held inside the product.
The old chips fizz again, sharing the work with the new ones.
STO-019 · medium · application · seq 10 · authored (shortcut-fails family: larger amount IS limiting) · mcq4
9.0 mol of Fe and 8.0 mol of O₂ react by 3Fe + 2O₂ → Fe₃O₄. Which reasoning correctly identifies the limiting reactant?
Fe — using all 9.0 mol of Fe needs only 6.0 mol of O₂, and 8.0 mol is available, so the O₂ outlasts the Fe. KEY
O₂ — it is present in the smaller amount, and the smaller amount is always limiting.
O₂ — using all 8.0 mol of O₂ would need 12.0 mol of Fe, and only 9.0 mol is present, so the O₂ cannot be fully used.
Neither — 9.0 mol against 8.0 mol is more O₂ than the equation could ever use.
STO-020 · medium · application · seq 11 · authored (routine branch: bigger pile limiting; galena roasting replaces 2ZnS+3O₂ — 'o 2 2 zno' token collision with 17A) · mcq4
4.0 mol of PbS and 5.0 mol of O₂ react by 2PbS + 3O₂ → 2PbO + 2SO₂. Which reactant is the limiting reactant?
O₂ KEY
PbS
PbO
Neither reactant
STO-020 · medium · application · seq 12 · authored (exact-ratio case at large coefficients; octane combustion; ledgered numbers 75.0/25/16/18 printed at this seq only) · mcq4
6.0 mol of C₈H₁₈ and 75.0 mol of O₂ burn by 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O. Which statement about the limiting reactant is correct?
Neither — the two reactants run out together. KEY
O₂ is the limiting reactant.
C₈H₁₈ is the limiting reactant.
It cannot be decided from the moles alone — the masses of the two reactants are needed as well.
STO-019 · medium · transfer · seq 13 · authored (claim class: smaller-mass rule — bridges toward STO-022's mass chain; instance-free) · mcq4
A revision card says: 'The reactant with the smaller MASS in grams is the limiting reactant.' Which reply is correct?
Wrong twice — grams must become moles first, and even those must be scaled by the coefficients. KEY
Right — less mass means less substance, and less substance runs out first.
Right, but only for reactions between two solids, where gas volumes can never confuse the comparison.
Wrong — the reactant with the LARGER mass is the limiting one.
STO-018 · medium · transfer · seq 14 · authored (claim class: time-will-convert-the-leftover) · mcq4
A student suggests: 'Given enough days, the leftover excess reactant will slowly turn into product on its own.' Which reply is correct?
Wrong — however long it waits, the leftover cannot become product without the other reactant. KEY
Right — over long times, all starting materials become products.
Right, provided the container stays tightly sealed so that nothing can escape.
Wrong — the leftover instead breaks down into new reactants.
The figure shows a reactant mixture for the reaction 2Na + S → Na₂S. Which particle diagram shows the mixture after the reaction stops?
FIGURE (stimulus · stimulus) — see raw specraw spec
{
"option_format": "image",
"render_contract": "R2 image-XOR-text: options render as pictures ONLY; the 'text'/'key' placeholder strings are internal ids for validation and are never rendered and never used for aria (R23 — descriptions must not exceed what a sighted student sees).",
"figureRequired": true,
"figureShape": "stimulus",
"stimulus": {
"figureType": "particle-diagram",
"box": "four Na atoms (each one medium gray circle, unjoined) and two S atoms (each one small yellow circle, unjoined), mixed together, widely spaced, no Na touching an S",
"key": "Na = one medium gray circle · S = one small yellow circle",
"style": "clean textbook particle diagram, single labeled box"
},
"optionFigureSpecs": [
{
"option": "key",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two Na₂S units (each two medium gray circles joined to one small yellow circle), mixed together, widely spaced, no other particles",
"key": "Na₂S = two medium gray circles joined to one small yellow circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor1",
"figureSpec": {
"figureType": "particle-diagram",
"box": "one Na₂S unit, two Na atoms (each one medium gray circle), and one S atom (one small yellow circle), mixed together, widely spaced",
"key": "Na₂S = two medium gray circles joined to one small yellow circle · Na = one medium gray circle · S = one small yellow circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor2",
"figureSpec": {
"figureType": "particle-diagram",
"box": "four product units drawn with the wrong recipe: each one medium gray circle joined to one small yellow circle, no other particles, mixed together, widely spaced",
"key": "product drawn as one medium gray circle joined to one small yellow circle",
"style": "clean textbook particle diagram, single labeled box"
}
},
{
"option": "distractor3",
"figureSpec": {
"figureType": "particle-diagram",
"box": "two Na₂S units (each two medium gray circles joined to one small yellow circle) and two Na atoms (each one medium gray circle), mixed together, widely spaced",
"key": "Na₂S = two medium gray circles joined to one small yellow circle · Na = one medium gray circle",
"style": "clean textbook particle diagram, single labeled box"
}
}
]
}
Picture 1 KEY
Picture 2
Picture 3
Picture 4
STO-021 · medium · application · seq 16 · authored (thirds ratio; tungsten reduction — 'tungsten' word-laned to this seq) · numeric
Tungsten metal is made by WO₃ + 3H₂ → W + 3H₂O. H₂ is the limiting reactant with 6.0 mol available, and 4.0 mol of WO₃ is present. How many moles of W form from this mixture? Give your answer in moles to 2 significant figures.
Answer: 2.0 mol KEY
also accepted: 2.0 mol
wrongValueDiagnostics
[
{
"value": 4.0,
"misconception": "you started from the excess reactant WO₃ — the limiting reactant H₂ caps the product",
"feedback": null
},
{
"value": 18.0,
"misconception": "you flipped the mole ratio — the coefficient of W goes on top and H₂'s underneath",
"feedback": null
},
{
"value": 6.0,
"misconception": "you carried the limiting reactant's moles straight across without the mole ratio",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of H₂ (limiting) = 6.0 mol\nThe 4.0 mol of WO₃ is the excess reactant — it does not enter the calculation.\nWrite down the equation:\nmoles of W = moles of H₂ × (coefficient of W ÷ coefficient of H₂)\nSubstitute in the values, and calculate:\nmoles of W = 6.0 × (1 ÷ 3)\nmoles of W = 2.0 mol"
}
STO-016 · medium · transfer · seq 17 · authored (bigger-pile-limiting family; ammonia oxidation without catalyst replaces ethyne — '2 c 2 h' token collision with 11A's ethane; equation distinct from 11B's 4NH₃ + 5O₂ Ostwald route) · mcq4
The figure shows a reactant mixture for the reaction 4NH₃ + 3O₂ → 2N₂ + 6H₂O. Which reactant is the limiting reactant?
FIGURE (particle-diagram · stimulus) — box: "eight NH₃ molecules (each one medium blue circle joined to three small light circles) and seven O₂ molecules (each two large red circles joined), mixed toge… · key: "NH₃ = one medium blue circle joined to three small light circles · O₂ = two large red circles joined" · style: "clean textbook particle diagram, single labeled box"raw spec
{
"figureRequired": true,
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-diagram",
"box": "eight NH₃ molecules (each one medium blue circle joined to three small light circles) and seven O₂ molecules (each two large red circles joined), mixed together, widely spaced, no two different molecules touching",
"key": "NH₃ = one medium blue circle joined to three small light circles · O₂ = two large red circles joined",
"style": "clean textbook particle diagram, single labeled box"
}
}
NH₃ KEY
O₂
N₂
Neither reactant
STO-021 · hard · application · seq 18 · authored (excess-trap with a halving ratio) · numeric
Silicon nitride ceramic forms by 3Si + 2N₂ → Si₃N₄. N₂ is the limiting reactant with 4.0 mol available; the vessel also holds 9.0 mol of Si. How many moles of Si₃N₄ does the reaction produce? Give your answer in moles to 2 significant figures.
Answer: 2.0 mol KEY
also accepted: 2.0 mol
wrongValueDiagnostics
[
{
"value": 3.0,
"misconception": "you started from the excess reactant Si — the limiting reactant N₂ caps the product",
"feedback": null
},
{
"value": 8.0,
"misconception": "you flipped the mole ratio — the coefficient of Si₃N₄ goes on top and N₂'s underneath",
"feedback": null
},
{
"value": 4.0,
"misconception": "you carried the limiting reactant's moles straight across without the mole ratio",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nmoles of N₂ (limiting) = 4.0 mol\nThe 9.0 mol of Si is the excess reactant — it does not enter the calculation.\nWrite down the equation:\nmoles of Si₃N₄ = moles of N₂ × (coefficient of Si₃N₄ ÷ coefficient of N₂)\nSubstitute in the values, and calculate:\nmoles of Si₃N₄ = 4.0 × (1 ÷ 2)\nmoles of Si₃N₄ = 2.0 mol"
}
STO-022 · hard · application · seq 19 · authored (mass chain; three-coefficient ratio in the limiting check) · numeric
5.400 g of Al and 63.92 g of Br₂ react to give AlBr₃: 2Al + 3Br₂ → 2AlBr₃. Molar masses (exact): Al 27.0 g/mol, Br₂ 159.8 g/mol, AlBr₃ 266.7 g/mol. Give the mass of AlBr₃ that forms, in grams (4 significant figures).
Answer: 53.34 g KEY
also accepted: 53.34 g
wrongValueDiagnostics
[
{
"value": 71.12,
"misconception": "you calculated from the excess reactant Br₂ — using up 0.20 mol of Al needs only 0.30 mol of Br₂, and 0.40 mol is available, so Al is limiting",
"feedback": null
},
{
"value": 69.32,
"misconception": "you added the two reactant masses — the leftover excess Br₂ is not part of the product",
"feedback": null
},
{
"value": 5.4,
"misconception": "you used Al's molar mass in the final step — the wanted substance AlBr₃'s molar mass is 266.7 g/mol",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm of Al = 5.400 g, M of Al = 27.0 g/mol, m of Br₂ = 63.92 g, M of Br₂ = 159.8 g/mol, M of AlBr₃ = 266.7 g/mol\nStep 1 — convert each mass to moles:\nn(Al) = 5.400 / 27.0 = 0.2000 mol\nn(Br₂) = 63.92 / 159.8 = 0.40 mol\nStep 2 — identify the limiting reactant:\nmoles of Br₂ needed = 0.2000 × (3 ÷ 2) = 0.3000 mol — 0.40 mol is available, so Al is limiting\nStep 3 — moles of product from the limiting reactant:\nmoles of AlBr₃ = 0.2000 × (2 ÷ 2) = 0.2000 mol\nStep 4 — convert the product moles to mass:\nm(AlBr₃) = 0.2000 × 266.7\nm(AlBr₃) = 53.34 g"
}
A problem states the mass of each of two reactants and asks what mass of one product forms. A student says: 'This is just a mass-to-mass problem — start from either reactant, whichever is easier.' Which reply is correct?
Wrong — the solver must first find the limiting reactant; a chain started from a reactant in excess overstates the product mass. KEY
Right — mass-to-mass chains give the same answer from either starting reactant.
Right, provided the answer is rounded to the correct number of significant figures at the end.
Wrong — with two reactant masses given, the problem cannot be solved without also knowing the reaction's temperature.
STO-020 · medium · application · seq 21 · authored (FRQ-1 form E: needed < available; 2.5 appears only in the unscanned model answer/rubric) · frq
A sealed vessel holds 5.0 mol of Ba and 4.0 mol of O₂. The reaction is 2Ba + O₂ → 2BaO.
A sealed vessel holds 5.0 mol of Ba and 4.0 mol of O₂. The reaction is 2Ba + O₂ → 2BaO.
(a) Calculate how many moles of O₂ would be needed to react completely with all 5.0 mol of Ba. Show your working.
(b) Compare the moles of O₂ needed with the moles of O₂ available, and say which reactant is limiting.
(c) Name the excess reactant, and state what will be true of it when the reaction stops.
Part a. Calculate how many moles of O₂ would be needed to react completely with all 5.0 mol of Ba. Show your working.
Model: Amounts supplied: 5.0 mol of barium, 4.0 mol of oxygen.
moles of oxygen needed = moles of barium × (coefficient of oxygen ÷ coefficient of barium)
moles of oxygen needed = 5.0 × (1 ÷ 2)
moles of oxygen needed = 2.5 mol
Part b. Compare the moles of O₂ needed with the moles of O₂ available, and say which reactant is limiting.
Model: 2.5 mol needed < 4.0 mol available — the oxygen will not run out, so the barium is used up first. The barium is the limiting reactant.
Part c. Name the excess reactant, and state what will be true of it when the reaction stops.
Model: The excess reactant is the oxygen. When the reaction stops, some of it will be left over, unreacted — it reacted only while the barium lasted.
Full model answer. (a) Amounts supplied: 5.0 mol of barium, 4.0 mol of oxygen.
moles of oxygen needed = moles of barium × (coefficient of oxygen ÷ coefficient of barium)
moles of oxygen needed = 5.0 × (1 ÷ 2)
moles of oxygen needed = 2.5 mol
(b) 2.5 mol needed < 4.0 mol available — the oxygen will not run out, so the barium is used up first. The barium is the limiting reactant.
(c) The excess reactant is the oxygen. When the reaction stops, some of it will be left over, unreacted — it reacted only while the barium lasted.
STO-022 · hard · application · seq 22 · authored (FRQ-2 form E: the tiny-mass reactant IS limiting) · frq
A chemist passes 28.05 g of C₄H₈ and 0.8000 g of H₂ over a catalyst; the reaction is C₄H₈ + H₂ → C₄H₁₀.
A chemist passes 28.05 g of C₄H₈ and 0.8000 g of H₂ over a catalyst; the reaction is C₄H₈ + H₂ → C₄H₁₀.
(a) Calculate the moles of each reactant.
(b) Identify the limiting reactant, showing the comparison of needed moles against available moles.
(c) Calculate the mass of C₄H₁₀ that forms. Show every step of your working.
Molar masses (exact): C₄H₈ 56.1 g/mol, H₂ 2.0 g/mol, C₄H₁₀ 58.1 g/mol.
Part b. Identify the limiting reactant, showing the comparison of needed moles against available moles.
Model: moles of hydrogen needed = moles of butene × (coefficient ratio) = 0.50 mol
0.50 mol needed > 0.40 mol available — the hydrogen falls short and runs out first: it is the limiting reactant.
Part c. Calculate the mass of C₄H₁₀ that forms. Show every step of your working.
Molar masses (exact): C₄H₈ 56.1 g/mol, H₂ 2.0 g/mol, C₄H₁₀ 58.1 g/mol.
Model: moles of butane = 0.4000 × (1 ÷ 1) = 0.4000 mol (starting from the limiting reactant)
mass of butane = 0.4000 × 58.1
mass of butane = 23.24 g
Full model answer. (a) moles of butene = 28.05 / 56.1 = 0.50 mol
moles of hydrogen = 0.8000 / 2.0 = 0.4000 mol
(b) moles of hydrogen needed = moles of butene × (coefficient ratio) = 0.50 mol
0.50 mol needed > 0.40 mol available — the hydrogen falls short and runs out first: it is the limiting reactant.
(c) moles of butane = 0.4000 × (1 ÷ 1) = 0.4000 mol (starting from the limiting reactant)
mass of butane = 0.4000 × 58.1
mass of butane = 23.24 g
End of Topic Test: Percent yield & Solution stoichiometry (u08-t3 · 5 forms × 22 items · 110 total)
David's notes for James:
CALC-HEAVY FRQ SPLIT - frq_per_form = 2, OVERRIDING your gate data's frq_per_form=1 baseline (confirm the composition; recorded adjudication A-U8G3-7): QB-v2 sec.4 directs 2-3 FRQs for calculation-heavy topics and names this exact class ('stoichiometry ... molarity'), and 4 of the 8 scope lessons are atomKind quant (STO-026/027/029/030) with the unit capstone's chain in scope. Count = one FRQ per DISTINCT taught multi-step procedure: FRQ-1 at seq 21 = STO-030's solution-stoichiometry chain (M x V -> moles -> mole ratio -> product mass; medium/application); FRQ-2 at seq 22 = STO-029's experiment-data percent-yield capstone (hard/application; theoretical yield from reactant data -> percent yield). A THIRD FRQ was considered and REJECTED: STO-026/027 are single-substitution skills fully measured by their numeric-ENTRY rows (seq 8/10/11/14/16). Total 20 objective + 2 FRQ = 22 items/form (inside the ~21-23 calc-heavy window); D5 scaling NOT triggered (8 lessons -> min(20, 48) = 20; your target_size field says 20 and agrees).
TAUGHT-CONSTANT '100' ADJUDICATION (TC-100, A-U8G3-6/A-U8G3-11) - THE HEADLINE FLAG for you AND the harness owner: the literal '100' is a taught constant of this whole scope (the 'x 100' step in every lesson's rule lines, plus the 100% believability cap that IS STO-028's content), but the shared validator's V-T7 numeric scan has no relief hook, so every item that states the equation or the cap fails against all six corpora as authored. Resolution: a RECORDED, fully-enumerated adjudicated instance family (25 slots = rows 3/5/7/12/19 x A-E; tokens ['100']; carrier vocabulary = the taught equation/cap discussion words) enforced by the gate-local validate_with_adjudications.py wrapper over the UNTOUCHED shared validator - exit 0, 388/388 fails excused, 0 uncovered, negative probes 5/5 (non-family '100' slot and '100.1' both stay hard fails; material-bearing and out-of-carrier '100' grams never excusable; CaCO3's molar mass 100.1 is NOT relieved - string-form exactness preserved). A proposed _shared strengthening (config-scoped taught_constant_numbers relief, default OFF, selftest probe pair) is DOCUMENTED NOT APPLIED because _shared was read-only this wave - please confirm the adjudication and the proposed generic mechanism.
D6 SCOPE GUARD (A-U8G3-9, confirm): the seq-22 FRQ part (a) asks the theoretical yield FROM REACTANT DATA - STO-029's taught steps 1-3. The D6 exclusion (STO-027 extraNotes) bans only the third permutation of the percent-yield equation (theoretical from actual + percent); no item on any form asks that permutation, and part (a) is the taught routine, not the banned one.
Title (A-U8G3-8): '(light)' stripped from Topic 5's manifest title as authoring metadata -> student-facing title 'End of Topic Test: Percent yield & Solution stoichiometry' (Topic-N prefixes also stripped, house rule).
Seed provenance (confirm the dispositions): pool 24 = the 8 lessons' spec gateItems positionally (24/24 stem match after notation normalization); demo_curated 3 are pool duplicates (indices 18/19/20, all STO-029 - dual-cited). 19 of the 20 form-A objective slots carry your seeds (seq 3, the STO-028 cap-rule recall, authored fresh; forms B-E authored fresh); prod_seeds is empty, so all 10 FRQs are our authored parallels of the taught STO-030/STO-029 procedures. VERBATIM placements: seq 2 (pool[7]), seq 14 (pool[18] = demo[0]), seq 17 (pool[4]). Everything else was freshened under the strict SEED RULING (instances/values only; structures, key logic, misconception classes kept verbatim) - highlights: seq 9 Na2SO4 -> K2SO4 with new M/V (sodium is an own-lesson material); seq 15 reaction swapped + new mL/M; seq 13 full dataset re-valued with results-only options; seq 20 (demo[2]) zinc kept with collected masses re-valued (60/75 -> 37/85); seq 12 '90' -> 95-class; seq 6 rebuilt at blind review to state a complete burn (the original 'too little air' framing left incomplete-combustion defensible - an F2 hard fail, fixed on all forms with an evidence-scoped ask). Spares 5 with promotion caveats recorded (pool[13] is numeric-as-MCQ and must convert to ENTRY if promoted; pool[22]'s answer 0.10 is a bare power of ten and must be re-valued).
Task-idiom / convention whitelists (A-U8G3-1..5, confirm - all task phrasing, zero instance content; materials and numbers in every stem still scan in full): STO-024's spec-mandated single-scenario classify frame ('below the theoretical yield. Which cause fits?'); STO-029's spec-mandated excess statement ('burns in excess O2 by' - your specNotes require stating excess in every stem); the three answer-format boilerplate sentences (moles/grams/percentage to N significant figures); STO-026's canonical stem template ('g. The experiment collects'); STO-028's taught cap-rule wording ('a genuine percent yield cannot rise above 100%').
Precision conventions (confirm): molar masses are declared '(exact)' on quantitative items (U8-G2 adjudication precedent). FRQ-22 part (a) instructs a 3-sf theoretical-yield report with supporting-precision intermediates shown in the model answer; part (b) divides the unrounded chain value, a rounded-value percent is recorded acceptable, and the percent sign OR the word 'percent' both count (typed answers never require special symbols). Numeric acceptedAlternates are kept ONLY when their numeric token equals the display exactly; every anticipated wrong value carries a student-facing 'This answer is wrong - you ...' diagnostic.
Accepted advisory classes from blind review (recorded, no fix): (a) cross-item cueing (74 flags) - recall rows state principles later calc rows use, and rows 13/20 share the taught raw-mass-comparison misconception catalog by blueprint design; (b) option-length flags (13) where the deterministic V-T4 longest-key balance PASSES; (c) stem-option-echo (11) - diagnosis tasks name their category (the category IS the option). THREE sol minors survive as recorded style notes (non-blocking, keys unaffected): A#20's worked explanation shows intermediate theoretical yields at 4 digits (exact-division values); C#18's expected mass 333 g vs the 74%-datum's 2-sf convention (the ask is already expectation-hedged); D#19's 'extra amounts are simply left over' (only the beyond-stoichiometric excess remains). Also confirmed at blind review: A#16's answer 50.0% matches an article-thread value (stage-1 WATCH) - both key-blind reviewers committed 50.0% cold, so the value carries no leak.
Instance hygiene: classroom-safe chemistry throughout - zero fissile/weapons/dose framings bank-wide (re-swept at packaging, 0 hits). Cross-lesson-material WARN pairs stand adjudicated and were screened for memory-answerability at blind review (keys picked without lesson access): A#6 carbon, A#20 zinc, D#6/D#20 copper, E#20 sodium (taught-elsewhere class); form D shares copper at seqs 6/20 and nickel at seqs 9/14 (non-adjacent lanes); form C carries manganese at seqs 9 and 21 (different constructs).
CRASH-RESUME PROVENANCE (disclosure, no action needed): the authoring fleet was killed TWICE mid-run by laptop restarts - once during stage 2 (after 20 of 22 rows), and again after the stage-7 finalize but BEFORE packaging. Both resumes worked from DISK STATE with full audits recorded in fix_log.md: every gate JSON artifact json.loads-verified (25/25 at the first resume, 18/18 relied-on artifacts at this packaging resume, nothing truncated), the forms payload re-verified byte-identical to the validated forms, and the adjudication-aware validator re-run fresh at packaging (exit 0, 388/388 TC-100 excusals, 0 uncovered - content-identical to the pre-crash stage-7 run). QC itself ran uninterrupted between the two crashes: blind rounds r1-r8 (key picks 198/198, keyless 112/112 after hand review, 2 F2 hard fails -> 0), sol authoritative loops 1-3 to 0 FAILs (110/110 pass, 3 recorded honest survivors). This export was built from the crash-audited forms_final.json.
FRQ delivery: your `prod` draw-on-paper class is delivered as a typed diagram-PLAN task (state kinds/recipes/counts/spacing; rubric grades the same observables). Confirm this mapping (pending from the U01-G1 round-trip).
full notes file
# u08-t3 — notes/queries from David's side (U8-G3)
1. CALC-HEAVY FRQ SPLIT - frq_per_form = 2, OVERRIDING your gate data's frq_per_form=1 baseline (confirm the composition; recorded adjudication A-U8G3-7): QB-v2 sec.4 directs 2-3 FRQs for calculation-heavy topics and names this exact class ('stoichiometry ... molarity'), and 4 of the 8 scope lessons are atomKind quant (STO-026/027/029/030) with the unit capstone's chain in scope. Count = one FRQ per DISTINCT taught multi-step procedure: FRQ-1 at seq 21 = STO-030's solution-stoichiometry chain (M x V -> moles -> mole ratio -> product mass; medium/application); FRQ-2 at seq 22 = STO-029's experiment-data percent-yield capstone (hard/application; theoretical yield from reactant data -> percent yield). A THIRD FRQ was considered and REJECTED: STO-026/027 are single-substitution skills fully measured by their numeric-ENTRY rows (seq 8/10/11/14/16). Total 20 objective + 2 FRQ = 22 items/form (inside the ~21-23 calc-heavy window); D5 scaling NOT triggered (8 lessons -> min(20, 48) = 20; your target_size field says 20 and agrees).
2. TAUGHT-CONSTANT '100' ADJUDICATION (TC-100, A-U8G3-6/A-U8G3-11) - THE HEADLINE FLAG for you AND the harness owner: the literal '100' is a taught constant of this whole scope (the 'x 100' step in every lesson's rule lines, plus the 100% believability cap that IS STO-028's content), but the shared validator's V-T7 numeric scan has no relief hook, so every item that states the equation or the cap fails against all six corpora as authored. Resolution: a RECORDED, fully-enumerated adjudicated instance family (25 slots = rows 3/5/7/12/19 x A-E; tokens ['100']; carrier vocabulary = the taught equation/cap discussion words) enforced by the gate-local validate_with_adjudications.py wrapper over the UNTOUCHED shared validator - exit 0, 388/388 fails excused, 0 uncovered, negative probes 5/5 (non-family '100' slot and '100.1' both stay hard fails; material-bearing and out-of-carrier '100' grams never excusable; CaCO3's molar mass 100.1 is NOT relieved - string-form exactness preserved). A proposed _shared strengthening (config-scoped taught_constant_numbers relief, default OFF, selftest probe pair) is DOCUMENTED NOT APPLIED because _shared was read-only this wave - please confirm the adjudication and the proposed generic mechanism.
3. D6 SCOPE GUARD (A-U8G3-9, confirm): the seq-22 FRQ part (a) asks the theoretical yield FROM REACTANT DATA - STO-029's taught steps 1-3. The D6 exclusion (STO-027 extraNotes) bans only the third permutation of the percent-yield equation (theoretical from actual + percent); no item on any form asks that permutation, and part (a) is the taught routine, not the banned one.
4. Title (A-U8G3-8): '(light)' stripped from Topic 5's manifest title as authoring metadata -> student-facing title 'End of Topic Test: Percent yield & Solution stoichiometry' (Topic-N prefixes also stripped, house rule).
5. Seed provenance (confirm the dispositions): pool 24 = the 8 lessons' spec gateItems positionally (24/24 stem match after notation normalization); demo_curated 3 are pool duplicates (indices 18/19/20, all STO-029 - dual-cited). 19 of the 20 form-A objective slots carry your seeds (seq 3, the STO-028 cap-rule recall, authored fresh; forms B-E authored fresh); prod_seeds is empty, so all 10 FRQs are our authored parallels of the taught STO-030/STO-029 procedures. VERBATIM placements: seq 2 (pool[7]), seq 14 (pool[18] = demo[0]), seq 17 (pool[4]). Everything else was freshened under the strict SEED RULING (instances/values only; structures, key logic, misconception classes kept verbatim) - highlights: seq 9 Na2SO4 -> K2SO4 with new M/V (sodium is an own-lesson material); seq 15 reaction swapped + new mL/M; seq 13 full dataset re-valued with results-only options; seq 20 (demo[2]) zinc kept with collected masses re-valued (60/75 -> 37/85); seq 12 '90' -> 95-class; seq 6 rebuilt at blind review to state a complete burn (the original 'too little air' framing left incomplete-combustion defensible - an F2 hard fail, fixed on all forms with an evidence-scoped ask). Spares 5 with promotion caveats recorded (pool[13] is numeric-as-MCQ and must convert to ENTRY if promoted; pool[22]'s answer 0.10 is a bare power of ten and must be re-valued).
6. Task-idiom / convention whitelists (A-U8G3-1..5, confirm - all task phrasing, zero instance content; materials and numbers in every stem still scan in full): STO-024's spec-mandated single-scenario classify frame ('below the theoretical yield. Which cause fits?'); STO-029's spec-mandated excess statement ('burns in excess O2 by' - your specNotes require stating excess in every stem); the three answer-format boilerplate sentences (moles/grams/percentage to N significant figures); STO-026's canonical stem template ('g. The experiment collects'); STO-028's taught cap-rule wording ('a genuine percent yield cannot rise above 100%').
7. Precision conventions (confirm): molar masses are declared '(exact)' on quantitative items (U8-G2 adjudication precedent). FRQ-22 part (a) instructs a 3-sf theoretical-yield report with supporting-precision intermediates shown in the model answer; part (b) divides the unrounded chain value, a rounded-value percent is recorded acceptable, and the percent sign OR the word 'percent' both count (typed answers never require special symbols). Numeric acceptedAlternates are kept ONLY when their numeric token equals the display exactly; every anticipated wrong value carries a student-facing 'This answer is wrong - you ...' diagnostic.
8. Accepted advisory classes from blind review (recorded, no fix): (a) cross-item cueing (74 flags) - recall rows state principles later calc rows use, and rows 13/20 share the taught raw-mass-comparison misconception catalog by blueprint design; (b) option-length flags (13) where the deterministic V-T4 longest-key balance PASSES; (c) stem-option-echo (11) - diagnosis tasks name their category (the category IS the option). THREE sol minors survive as recorded style notes (non-blocking, keys unaffected): A#20's worked explanation shows intermediate theoretical yields at 4 digits (exact-division values); C#18's expected mass 333 g vs the 74%-datum's 2-sf convention (the ask is already expectation-hedged); D#19's 'extra amounts are simply left over' (only the beyond-stoichiometric excess remains). Also confirmed at blind review: A#16's answer 50.0% matches an article-thread value (stage-1 WATCH) - both key-blind reviewers committed 50.0% cold, so the value carries no leak.
9. Instance hygiene: classroom-safe chemistry throughout - zero fissile/weapons/dose framings bank-wide (re-swept at packaging, 0 hits). Cross-lesson-material WARN pairs stand adjudicated and were screened for memory-answerability at blind review (keys picked without lesson access): A#6 carbon, A#20 zinc, D#6/D#20 copper, E#20 sodium (taught-elsewhere class); form D shares copper at seqs 6/20 and nickel at seqs 9/14 (non-adjacent lanes); form C carries manganese at seqs 9 and 21 (different constructs).
10. CRASH-RESUME PROVENANCE (disclosure, no action needed): the authoring fleet was killed TWICE mid-run by laptop restarts - once during stage 2 (after 20 of 22 rows), and again after the stage-7 finalize but BEFORE packaging. Both resumes worked from DISK STATE with full audits recorded in fix_log.md: every gate JSON artifact json.loads-verified (25/25 at the first resume, 18/18 relied-on artifacts at this packaging resume, nothing truncated), the forms payload re-verified byte-identical to the validated forms, and the adjudication-aware validator re-run fresh at packaging (exit 0, 388/388 TC-100 excusals, 0 uncovered - content-identical to the pre-crash stage-7 run). QC itself ran uninterrupted between the two crashes: blind rounds r1-r8 (key picks 198/198, keyless 112/112 after hand review, 2 F2 hard fails -> 0), sol authoritative loops 1-3 to 0 FAILs (110/110 pass, 3 recorded honest survivors). This export was built from the crash-audited forms_final.json.
Form A
STO-023 · easy · recall · seq 1 · seed:U8-G3 pool[2] (STO-023) — value freshened 7.5 g -> 6.2 g ('7.5' in STO-026 WE2 corpus; '6.2' ledger-free); structure, key, and all three misconception classes kept verbatim; explanation swept for the old value · mcq4
Before running a reaction, a chemist calculates that 6.2 g of product can form. No product has been made yet. What is the 6.2 g?
The theoretical yield. KEY
The actual yield.
Both yields at once.
Neither — it becomes a yield only after the experiment.
STO-025 · easy · recall · seq 2 · seed:U8-G3 pool[7] (STO-025, James gateItem) — placed VERBATIM (instance-free, no numbers in stem/options, scan-clean per seed_placement); key authored at option a; explanation carried verbatim (its '× 100' lives on the unscanned feedback side) · mcq4
Which statement describes what a percent yield reports?
The measured product as a percentage of the predicted product. KEY
The predicted product as a percentage of the measured product.
The mass of product lost during the experiment.
The fraction of the limiting reactant that dissolved.
What is the highest value a genuine percent yield can reach?
Exactly 100% — the collected product can at most equal the calculated maximum. KEY
There is no fixed ceiling — unusually good technique can push a genuine yield past 100%.
Slightly above 100%, because normal experimental variation has to be allowed for.
Just below 100% — a genuine run can approach the maximum but never actually reach it.
STO-023 · easy · application · seq 4 · seed:U8-G3 pool[0] (STO-023 gate item) — VALUES freshened per SEED RULING: calculated entry 25 -> 26 g ('25' hits STO-026 rule instance + STO-027 corpus; placement example 23 g swapped to 26 g so the 4 g gap does not echo the seq-17 form-A seed's '4 g' shortfall); measured 19 g kept (ledger-free); difference distractor 6 -> 7 g re-derived; all fields swept (QB 3.9) · mcq4
A lab notebook for one reaction shows two entries: 'from calculation: 26 g' and 'collected and weighed: 19 g'. Which entry is the actual yield?
The 19 g, because it was measured after the experiment. KEY
The 26 g, because it is the more reliable number.
Both entries, because each describes the product.
Neither entry — the actual yield is the 7 g difference between them.
Two students write the percent yield equation from memory. Student 1 writes percent yield = (actual yield ÷ theoretical yield) × 100. Student 2 writes percent yield = (theoretical yield ÷ actual yield) × 100. Who is right?
Student 1 — the actual yield is divided by the theoretical yield. KEY
Student 2 — the theoretical yield is divided by the actual yield.
Both — the two versions give the same number.
Neither — the equation needs the two yields multiplied, not divided.
STO-024 · medium · application · seq 6 · seed:U8-G3 pool[3] (STO-024, James gateItem) — placed VERBATIM (Unicode transport only); key authored at option a; 'carbon' cross-lesson A5 WARN recorded (seed_placement seq 6); closing classify frame covered by gate_config extra_task_idioms[0]; explanation carried verbatim | blind-wave1 2026-09-08 (SEED RULING content fix): complete-burn stated (removes the too-little-air incomplete-combustion confound) + evidence-scoped ask (F2 A#6) · mcq4
A charcoal sample burns away completely, but some of its carbon forms carbon monoxide instead of carbon dioxide. A student collecting the CO₂ finds the actual yield below the theoretical yield. Which cause does this observation directly show?
A side reaction consumed reactant. KEY
Product was lost on transfer.
The reaction stopped before completion.
Atoms were destroyed in the flame.
STO-025 · medium · application · seq 7 · seed:U8-G3 pool[6] (STO-025, James gateItem) — VALUES freshened 36/9 -> 84/21 (36 = STO-027 article-register hit per seed_placement; the placement's example 42/14 rejected — 14 is form A seq 10's kept value; 84/21 preserves the seed's ÷4 clean-ratio class, both strings prescan-clean); structure, key at option a, all three distractor classes and the explanation pattern kept; every field swept for the old 36/9 (QB 3.9); options' literal '100' -> taught_constant_numbers adjudication (A-U8G3-6) · mcq4
A reaction's theoretical yield is 84 g, and its actual yield is 21 g. Which substitution into the percent yield equation is correct?
(21 ÷ 84), multiplied by 100 KEY
(84 ÷ 21), multiplied by 100
(21 ÷ 84), divided by 100
(84 − 21), read directly as a percentage
STO-026 · medium · application · seq 8 · seed:U8-G3 pool[9] (STO-026) — wording freshened per the must_freshen_wording flag: 'A pilot batch... The plant collects' -> 'A trial run of a reaction... The experiment collects' (cross-lesson gram 'batch has a theoretical' hits STO-027 fluency; template phrase = gate_config A-U8G3-4); values 22.5/25.0, 90.0% @3sf contract, scaffold, and flip/forgot-x100 diagnostics kept verbatim; subtracted diagnostic added per the specNotes fixed set; fields swept for batch/plant leftovers (QB 3.9) · numeric
A trial run of a reaction has a theoretical yield of 25.0 g. The experiment collects 22.5 g of product. What is the percent yield? Give your answer as a percentage to 3 significant figures.
Answer: 90.0 % KEY
also accepted: 90.0%
wrongValueDiagnostics
[
{
"value": 111.1,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 0.9,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
},
{
"value": 2.5,
"misconception": "you subtracted the yields — percent yield compares them by division, not by difference",
"feedback": "This answer is wrong — you subtracted the yields — percent yield compares them by division, not by difference."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nactual yield = 22.5 g\ntheoretical yield = 25.0 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nSubstitute in the values, and calculate:\npercent yield = (22.5 ÷ 25.0) × 100\npercent yield = 90.0%"
}
STO-030 · medium · application · seq 9 · seed:U8-G3.pool[21] (STO-030 gateItems[0], BaCl₂ precipitation) — excess partner swapped Na₂SO₄ → K₂SO₄ with product 2NaCl → 2KCl ('sodium' = own-lesson material FAIL); M/V freshened 0.25 L / 0.20 M → 0.35 L / 0.16 M ('0.25' STO-029 corpus, '0.20' own corpus); structure, 1:1 ratio, key route, and both misconception classes kept; every field swept for sodium/Na and the old values (QB 3.9) · numeric
0.35 L of 0.16 M barium chloride solution reacts with excess potassium sulfate by BaCl₂ + K₂SO₄ → BaSO₄ + 2KCl. How many moles of BaSO₄ form? Give your answer in moles to 2 significant figures.
Answer: 0.056 mol KEY
also accepted: 0.056 mol
wrongValueDiagnostics
[
{
"value": 2.2,
"misconception": "you divided instead of multiplying — n = M × V multiplies the molarity by the volume",
"feedback": "This answer is wrong — you divided instead of multiplying — n = M × V multiplies the molarity by the volume."
},
{
"value": 0.11,
"misconception": "you doubled the moles with KCl's coefficient — BaSO₄'s ratio to BaCl₂ is 1 : 1",
"feedback": "This answer is wrong — you doubled the moles with KCl's coefficient — BaSO₄'s ratio to BaCl₂ is 1 : 1."
}
]
feedback
{
"keyExplanation": "Step 1 — moles of dissolved reactant:\nn(BaCl₂) = M × V\nn(BaCl₂) = 0.16 × 0.35 = 0.056 mol\nStep 2 — moles of product with the mole ratio:\nmoles of BaSO₄ = 0.056 × (1 ÷ 1)\nmoles of BaSO₄ = 0.056 mol"
}
STO-026 · medium · application · seq 10 · seed:U8-G3 pool[10] (STO-026) — placed VERBATIM: value strings 14/56 ledger-free; template gram 'g. The experiment collects' + answer-format sentence covered by gate_config adjudications A-U8G3-4/A-U8G3-3; error-analysis narrative decomposed into the misconceptions list; scaffold + sanity-check framing kept · numeric
A reaction has a theoretical yield of 56 g. The experiment collects 14 g of product. What is the percent yield? Give your answer as a percentage to 3 significant figures.
Answer: 25.0 % KEY
also accepted: 25.0%
wrongValueDiagnostics
[
{
"value": 400,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 0.25,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
},
{
"value": 42,
"misconception": "you subtracted the actual yield from the theoretical yield and reported the difference as a percent",
"feedback": "This answer is wrong — you subtracted the actual yield from the theoretical yield and reported the difference as a percent."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nactual yield = 14 g\ntheoretical yield = 56 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nSubstitute in the values, and calculate:\npercent yield = (14 ÷ 56) × 100\npercent yield = 25.0%\nThe actual yield is far below the theoretical yield, so the percentage must come out well below 100% — an answer like 400% fails that check."
}
STO-027 · medium · application · seq 11 · seed:U8-G3 pool[12] (STO-027, James gateItem) — VALUES re-set 400 g/85% -> 175 g/44% per the V-T6 preferred resolution (seed display '340'@2sf = trailing-zero-ambiguous; no ledger-free theoretical keeps 85% with a zero-free 2-digit exact product); construct, production-run surface, question sentence, and all three diagnostic classes kept; every field swept for 400/85/340 residue (QB 3.9); tail = gate_config whitelisted convention phrase (A-U8G3-3) · numeric
A production run has a theoretical yield of 175 g, and the process reliably gives a 44% yield. What mass of product should the run expect to collect? Give your answer in grams to 2 significant figures.
Answer: 77 g KEY
also accepted: 77 g
wrongValueDiagnostics
[
{
"value": 397.7,
"misconception": "you divided the theoretical yield by the percent fraction instead of multiplying — a below-100% yield must give an actual yield below the theoretical 175 g",
"feedback": "This answer is wrong — you divided the theoretical yield by the percent fraction instead of multiplying — a below-100% yield must give an actual yield below the theoretical 175 g."
},
{
"value": 7700,
"misconception": "you forgot to divide the percent yield by 100 — thousands of grams from a 175 g prediction is the signal",
"feedback": "This answer is wrong — you forgot to divide the percent yield by 100 — thousands of grams from a 175 g prediction is the signal."
},
{
"value": 131,
"misconception": "you subtracted the percent number from the theoretical yield as if it were a mass in grams — 44 is a percentage, not a mass",
"feedback": "This answer is wrong — you subtracted the percent number from the theoretical yield as if it were a mass in grams — 44 is a percentage, not a mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\npercent yield = 44%\ntheoretical yield = 175 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nMake the actual yield the subject:\nactual yield = (percent yield ÷ 100) × theoretical yield\nSubstitute in the values, and calculate:\nactual yield = (44 ÷ 100) × 175\nactual yield = 77 g\nSanity check: 77 g is below the theoretical 175 g — a percent yield below 100% must give an actual yield below the theoretical yield."
}
STO-028 · medium · application · seq 12 · seed:U8-G3 pool[15] (STO-028, James gateItem) — option-c value freshened 90 -> 95 ('90' = STO-027/STO-029 corpus hit per seed_placement; '95' prescan-clean; invented-cap coherence kept, 95 < 96); all else verbatim incl. 96/103 (ledger-free; 103 = specNotes over-cap floor) and distractor-d's 110; every field swept for the old 90 (QB 3.9); key's 'a genuine percent yield cannot rise above 100%' covered by gate_config A-U8G3-5; literal '100' -> taught_constant_numbers adjudication (A-U8G3-6) · mcq4
Two groups ran the same synthesis. Group A reports a 96% yield; Group B reports a 103% yield. Which report needs rechecking?
Group B's — a genuine percent yield cannot rise above 100%. KEY
Group A's — 96% is too close to perfect to be honest.
Both reports — yields above 95% are impossible in real labs.
Neither report — values up to 110% are normal experimental variation.
STO-026 · medium · application · seq 13 · seed:U8-G3 pool[11] (STO-026 gate item) — FULL DATASET RE-VALUE per seed_placement (20/15/50/30 -> 72/63/320/136; results 75/60 -> 87.5/42.5; flipped pair 167/133 -> 235/114; old values hit four lessons' corpora incl. own and 75/60 are register results); structure, key at option a, winner-first-with-fewer-grams inversion, and all three distractor classes kept; every field swept for old values (QB 3.9); options results-only — no '100' in stem/options · mcq4
Experiment A has a theoretical yield of 72 g and collects 63 g. Experiment B has a theoretical yield of 320 g and collects 136 g. Which experiment has the higher percent yield?
Experiment A — 87.5% compared with 42.5%. KEY
Experiment B — collecting more grams of product always means the higher percent yield.
Experiment B — 235% compared with 114%.
The two experiments tie — falling short of a prediction counts the same however big the shortfall is.
STO-029 · medium · application · seq 14 · seed:U8-G3 pool[18] = demo_curated[0] (STO-029, James gateItem/demo-gate) — placed VERBATIM per seed_placement (scan-clean; CaCO₃ family corpus-excluded, reserved to form A); full seed contract carried (80.0 / ±0.05 / % / 3 sf); explanation_correct verbatim; explanation_incorrect's two diagnostics (44.8, 125) carried into the misconceptions schema · numeric
50.05 g of CaCO₃ is heated and decomposes by CaCO₃ → CaO + CO₂. 22.44 g of CaO is collected. What is the percent yield of CaO? (M of CaCO₃ = 100.1 g/mol; M of CaO = 56.1 g/mol.) Give your answer as a percentage to 3 significant figures.
Answer: 80.0 % KEY
also accepted: 80.0%
wrongValueDiagnostics
[
{
"value": 44.8,
"misconception": "you divided the product mass by the reactant mass — the comparison is with the theoretical yield, 28.05 g",
"feedback": "This answer is wrong — you divided the product mass by the reactant mass — the comparison is with the theoretical yield, 28.05 g."
},
{
"value": 125,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
}
]
feedback
{
"keyExplanation": "Step 1 — convert the reactant mass to moles:\nn(CaCO₃) = 50.05 / 100.1 = 0.500 mol\nStep 2 — convert to moles of product with the mole ratio:\nmoles of CaO = 0.500 × (1 ÷ 1) = 0.500 mol\nStep 3 — convert the product moles to mass (the theoretical yield):\ntheoretical yield = 0.500 × 56.1 = 28.05 g\nStep 4 — calculate the percent yield:\npercent yield = (22.44 ÷ 28.05) × 100\npercent yield = 80.0%"
}
STO-030 · hard · application · seq 15 · seed:U8-G3.pool[23] (STO-030 gateItems[2], the mL-trap gate item) — reaction swapped Mg + CuCl₂ → SrCl₂ + K₂SO₄ → SrSO₄ + 2KCl per must_freshen_instance ('copper' + 'magnesium' = own-lesson material FAILs; SrCl₂ + K₂SO₄ is the blueprint/seed_placement candidate, and row09's coordination note reserved strontium to this row and recorded the shared K₂SO₄ form-A partner as blueprint-sanctioned — different seq, no A5 class); M/V freshened 250 mL / 0.80 M → 750 mL / 0.32 M per must_freshen_values ('250' STO-027 corpus, '0.80' own corpus); answer re-derived 0.20 → 0.24 mol; mL→L conversion lead, 1:1 ratio, all three diagnostic classes, scaffold shape, and the sanity-check line kept; every field swept for copper/Cu/magnesium/Mg/CuCl₂/MgCl₂ and the old values 250/0.250/0.80/0.20/200/0.31 (QB 3.9) | 2026-09-08 crash-resume: M re-freshened 0.28 → 0.32 ('0.28' collided cross-form with the B#21 FRQ prompt), answer re-derived 0.24 mol; diagnostics and scaffold re-derived. · numeric
750 mL of 0.32 M strontium chloride solution is combined with excess potassium sulfate by SrCl₂ + K₂SO₄ → 2KCl + SrSO₄. How many moles of SrSO₄ form? Give your answer in moles to 2 significant figures.
Answer: 0.24 mol KEY
also accepted: 0.24 mol
wrongValueDiagnostics
[
{
"value": 240.0,
"misconception": "you used the volume in milliliters — n = M × V needs liters, and 750 mL is 0.750 L",
"feedback": "This answer is wrong — you used the volume in milliliters — n = M × V needs liters, and 750 mL is 0.750 L."
},
{
"value": 0.32,
"misconception": "you used the molarity as the mole amount without multiplying by the volume",
"feedback": "This answer is wrong — you used the molarity as the mole amount without multiplying by the volume."
},
{
"value": 2.34,
"misconception": "you divided the volume by the molarity instead of multiplying them",
"feedback": "This answer is wrong — you divided the volume by the molarity instead of multiplying them."
}
]
feedback
{
"keyExplanation": "Convert the volume to liters first: 750 mL = 0.750 L.\nStep 1 — moles of dissolved reactant:\nn(SrCl₂) = M × V\nn(SrCl₂) = 0.32 × 0.750 = 0.24 mol\nStep 2 — moles of product with the mole ratio:\nmoles of SrSO₄ = 0.24 × (1 ÷ 1)\nmoles of SrSO₄ = 0.24 mol\nA 240 mol answer fails the sanity check — three-quarters of a liter of a dilute solution holds well under a mole of solute."
}
STO-029 · hard · application · seq 16 · seed:U8-G3 pool[19] (STO-029) VERBATIM — james_curated, == demo_curated[1] (dual citation); notation transported (sub-tag markup / HTML arrow entities) -> Unicode per the James-export mandate; prescan idiom-only ('burns in excess O₂ by' = gate_config A-U8G3-2); diagnostics lifted verbatim from the seed's explanation_incorrect; WATCH: answer 50.0% equals the article thread's 50% (different reaction) — blind round to confirm no article-recall answerability · numeric
12.84 g of sulfur burns in excess O₂ by S + O₂ → SO₂, and 12.82 g of SO₂ is collected. What is the percent yield? (M of S = 32.1 g/mol; M of SO₂ = 64.1 g/mol.) Give your answer as a percentage to 3 significant figures.
Answer: 50.0 % KEY
also accepted: 50.0%
wrongValueDiagnostics
[
{
"value": 0.5,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
},
{
"value": 99.8,
"misconception": "you compared the product mass with the reactant mass — the theoretical yield is 25.64 g of SO₂, not 12.84 g",
"feedback": "This answer is wrong — you compared the product mass with the reactant mass — the theoretical yield is 25.64 g of SO₂, not 12.84 g."
},
{
"value": 200.0,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
}
]
feedback
{
"keyExplanation": "Step 1 — convert the reactant mass to moles:\nn(S) = 12.84 / 32.1 = 0.400 mol\nStep 2 — convert to moles of product with the mole ratio:\nmoles of SO₂ = 0.400 × (1 ÷ 1) = 0.400 mol\nStep 3 — convert the product moles to mass (the theoretical yield):\ntheoretical yield = 0.400 × 64.1 = 25.64 g\nStep 4 — calculate the percent yield:\npercent yield = (12.82 ÷ 25.64) × 100\npercent yield = 50.0%\nThe near-match between 12.82 g of product and 12.84 g of reactant is a coincidence of masses — the yield comparison is always against the theoretical yield."
}
STO-024 · hard · transfer · seq 17 · seed:U8-G3 pool[4] (STO-024, James gateItem) — placed VERBATIM (no notation transport needed); key authored at option a; scan-clean per seed_placement seq 17 (4 <= 12 ignored, no materials, conservation option wording whitelist-derived); seed explanation carried verbatim · mcq4
A student's actual yield is 4 g below the theoretical yield. A classmate says the missing 4 g proves that atoms were destroyed during the reaction. What is the correct account?
No atoms were destroyed — the missing mass is matter that ended up elsewhere, such as unreacted material, side products, or residue on equipment. KEY
The classmate is right — every chemical reaction destroys a small fraction of its atoms.
The missing mass was converted into the heat that the reaction gave off as it ran, so a warmer reaction always weighs out lighter.
The theoretical yield counted some of the atoms twice, so the prediction came out too high.
STO-027 · hard · transfer · seq 18 · seed:U8-G3 pool[14] (STO-027, James gateItem) — full value set freshened 30/60/45/27/33 -> 102/225/44/99/126 per seed_placement must_freshen_values ('30' STO-026 corpus, '60' STO-027+STO-029 corpora, '45'/'27' STO-027 corpus, '33' burned in rows 04/07/10; replacements prescan-clean); No verdict and 3 g shortfall margin kept; flipped-percentage feedback reworded numeral-free (no literal '100' anywhere — results-only advisory honored); structure, key position, and distractor classes verbatim-in-kind; all fields swept for old values (QB 3.9) · mcq4
A customer orders 102 g of a compound. The planned run has a theoretical yield of 225 g, and the reaction typically gives a 44% yield. Is the run expected to deliver the order?
No — the expected actual yield is 99 g, below the 102 g ordered. KEY
Yes — a theoretical yield of 225 g guarantees the run collects at least the 102 g ordered.
Yes — the expected actual yield is 126 g, comfortably above the 102 g ordered.
It cannot be judged — an expected mass cannot be calculated without the molar masses of the compound and reactants.
A student argues: 'With enough excess reactant, a percent yield above 100% is possible, because the extra reactant makes extra product.' What is wrong with the argument?
The theoretical yield already counts all of the limiting reactant — excess cannot push the product above it. KEY
Nothing is wrong — with plenty of excess reactant, real runs do climb above 100%.
The argument fails only because a large excess of reactant slows the reaction down.
The argument fails because adding excess reactant lowers the theoretical yield instead.
STO-029 · hard · transfer · seq 20 · seed:U8-G3 pool[20] (STO-029 gate item) = demo_curated[2] (dual citation; the demo version differs only by distractor order) — COLLECTED masses re-valued per must_freshen_values (24.42/12.21 g → 15.059/13.838 g): the seed's results 60/75 sit in STO-029's own fluency corpus; new results 37% and 85% are spec- and inventory-clean, and the flipped pair re-derives 167/133 → 270/118. Reactant masses 32.70/13.08 g and both molar masses kept (real Zn/ZnO data, exact division, gateItem text is corpus-excluded). Run 2's collected 13.838 g now sits just above its own reactant mass class — the tempting raw-grams comparison (15.059 > 13.838) still points at the loser. 'zinc' = cross-lesson WARN (STO-030 rule example Zn + HCl), recorded at stage 1. · mcq4
Two runs of 2Zn + O₂ → 2ZnO burn different amounts of zinc in excess O₂ (M of Zn = 65.4 g/mol; M of ZnO = 81.4 g/mol). Run 1: 32.70 g of Zn, 15.059 g of ZnO collected. Run 2: 13.08 g of Zn, 13.838 g of ZnO collected. Which run has the higher percent yield?
Run 2 — 85% compared with 37%. KEY
Run 1 — collecting more grams of ZnO (15.059 g against 13.838 g) must mean the higher percent yield.
Run 1 — 270% compared with 118%.
The runs tie — the same reaction always gives the same percent yield, whatever the size of the load.
STO-030 · medium · application · seq 21 · authored (FRQ-1 form A: fresh taught-class precipitation Mn(NO₃)₂ + 2KOH → Mn(OH)₂ + 2KNO₃, 1 : 1; manganese is absent from every scope corpus and sits in a different form than seq 9's manganese; liters given directly; exact product 0.0798 mol → 0.080 mol at 2 s.f.) · frq
0.38 L of 0.21 M aqueous manganese(II) nitrate is treated with excess potassium hydroxide by Mn(NO₃)₂ + 2KOH → Mn(OH)₂ + 2KNO₃.
0.38 L of 0.21 M aqueous manganese(II) nitrate is treated with excess potassium hydroxide by Mn(NO₃)₂ + 2KOH → Mn(OH)₂ + 2KNO₃.
(a) Calculate how many moles of Mn(NO₃)₂ are dissolved in the 0.38 L of solution. Show your working.
(b) Use the mole ratio from the balanced equation to calculate how many moles of Mn(OH)₂ form. Show your working. Give your answer in moles to 2 significant figures.
Part a. Calculate how many moles of Mn(NO₃)₂ are dissolved in the 0.38 L of solution. Show your working.
Model: Values: V = 0.38 L of solution, M = 0.21 mol/L — M is the molarity of the manganese(II) nitrate solution, not a molar mass.
n = M × V
n = 0.21 × 0.38
n = 0.0798 mol of manganese(II) nitrate dissolved.
Part b. Use the mole ratio from the balanced equation to calculate how many moles of Mn(OH)₂ form. Show your working. Give your answer in moles to 2 significant figures.
Model: From the balanced equation, 1 mol of manganese(II) hydroxide forms for every 1 mol of manganese(II) nitrate.
moles of manganese(II) hydroxide = 0.0798 × (1 ÷ 1)
moles of manganese(II) hydroxide = 0.0798 mol
To 2 significant figures: 0.080 mol of manganese(II) hydroxide. The product stays in moles.
Full model answer. (a) Values: V = 0.38 L of solution, M = 0.21 mol/L — M is the molarity of the manganese(II) nitrate solution, not a molar mass.
n = M × V
n = 0.21 × 0.38
n = 0.0798 mol of manganese(II) nitrate dissolved.
(b) From the balanced equation, 1 mol of manganese(II) hydroxide forms for every 1 mol of manganese(II) nitrate.
moles of manganese(II) hydroxide = 0.0798 × (1 ÷ 1)
moles of manganese(II) hydroxide = 0.0798 mol
To 2 significant figures: 0.080 mol of manganese(II) hydroxide. The product stays in moles.
A student ignites 91.9 g of tungsten powder, which burns in excess O₂ by 2W + 3O₂ → 2WO₃. After cooling, 81.13 g of WO₃ is collected.
A student ignites 91.9 g of tungsten powder, which burns in excess O₂ by 2W + 3O₂ → 2WO₃. After cooling, 81.13 g of WO₃ is collected.
(a) Calculate the theoretical yield of WO₃. Show every step of your working. Report the theoretical yield to 3 significant figures.
(b) Calculate the percent yield of WO₃. Give your answer as a percentage to 3 significant figures.
Molar masses (exact): W 183.8 g/mol, WO₃ 231.8 g/mol.
Write your answers in plain numbers and words; no special symbols are needed.
Part a. Calculate the theoretical yield of WO₃. Show every step of your working. Report the theoretical yield to 3 significant figures.
Model: Values: mass of tungsten = 91.9 g; molar mass of tungsten = 183.8 g/mol; molar mass of tungsten trioxide = 231.8 g/mol.
Equation: n = m ÷ M
Substitute: n = 91.9 ÷ 183.8
Calculate: n = 0.500 mol of tungsten
Mole ratio from the balanced equation: 2 mol of tungsten trioxide forms from 2 mol of tungsten, so n = 0.500 × (2 ÷ 2) = 0.500 mol of tungsten trioxide
Equation: m = n × M
Substitute: m = 0.500 × 231.8
Calculate: theoretical yield = 115.90 g, reported as 116 g to 3 significant figures
Part b. Calculate the percent yield of WO₃. Give your answer as a percentage to 3 significant figures.
Molar masses (exact): W 183.8 g/mol, WO₃ 231.8 g/mol.
Write your answers in plain numbers and words; no special symbols are needed.
Model: Values: actual yield = 81.13 g; theoretical yield = 115.90 g.
Equation: percent yield = (actual yield ÷ theoretical yield) × 100
Substitute: percent yield = (81.13 ÷ 115.90) × 100
Calculate: percent yield = 70.0% (dividing by the rounded 116 g gives 69.9%, which also earns full credit)
Full model answer. (a) Values: mass of tungsten = 91.9 g; molar mass of tungsten = 183.8 g/mol; molar mass of tungsten trioxide = 231.8 g/mol.
Equation: n = m ÷ M
Substitute: n = 91.9 ÷ 183.8
Calculate: n = 0.500 mol of tungsten
Mole ratio from the balanced equation: 2 mol of tungsten trioxide forms from 2 mol of tungsten, so n = 0.500 × (2 ÷ 2) = 0.500 mol of tungsten trioxide
Equation: m = n × M
Substitute: m = 0.500 × 231.8
Calculate: theoretical yield = 115.90 g, reported as 116 g to 3 significant figures
(b) Values: actual yield = 81.13 g; theoretical yield = 115.90 g.
Equation: percent yield = (actual yield ÷ theoretical yield) × 100
Substitute: percent yield = (81.13 ÷ 115.90) × 100
Calculate: percent yield = 70.0% (dividing by the rounded 116 g gives 69.9%, which also earns full credit)
The theoretical yield is the most product the reactants can possibly make, and the actual yield can never genuinely rise above that maximum. KEY
Because chemists agree to round any higher measured value down to 100% when reporting.
Because the final ×100 step of the equation cannot produce an answer bigger than 100.
Because the theoretical yield is deliberately calculated on the high side, so no real experimental run can ever quite catch up to it.
STO-023 · easy · application · seq 4 · authored (surface: batch record; probe: theoretical yield; trust class inverted for the theoretical probe; values 38/33 ledger-verified) · mcq4
A batch record for one reaction lists two masses: 'calculated before the run: 38 g' and 'weighed at the end of the run: 33 g'. Which mass is the theoretical yield?
The 38 g, because it comes from the stoichiometry calculation. KEY
The 33 g, because a weighed mass is more dependable than a calculated one.
Both masses, because each one describes the product of the run.
Neither mass — the theoretical yield is the 5 g difference between them.
Two lab partners write the percent yield equation on a whiteboard from memory. Partner 1 writes percent yield = (theoretical yield ÷ actual yield) × 100. Partner 2 writes percent yield = (actual yield ÷ theoretical yield) × 100. Who is right?
Partner 2 — the actual yield is divided by the theoretical yield. KEY
Partner 1 — the theoretical yield is divided by the actual yield.
Both — a fraction gives the same value whichever yield goes on top.
Neither — the equation should subtract the actual yield from the theoretical yield.
STO-024 · medium · application · seq 6 · authored (transfer-loss rotation: filter-funnel residue surface per blueprint; taught bare cause labels constant, anchor distractor scenario-fitted; 'filter paper' avoided — funnel/weighing-bottle surface keeps the stem gram-distinct from WE1 and fluency-2) | blind-wave1 2026-09-08: evidence-scoped ask (F2 class sweep) · mcq4
A student dries a solid product in a filter funnel, then tips it into a weighing bottle. A film of powder stays smeared inside the funnel and cannot be tipped out. The actual yield lands below the theoretical yield. Which cause does this observation directly show?
Product was lost on transfer. KEY
A side reaction consumed reactant.
The reaction stopped before completion.
Atoms in the powder were destroyed as it dried.
STO-025 · medium · application · seq 7 · authored (fresh exact-division yield pair 55/33, ratio 3:5, both strings prescan-clean; seed p6's structure and distractor classes kept; 'has a theoretical yield of' = canonical-lexicon subset gram, the derivation that clears seq 10's seed) · mcq4
A synthesis has a theoretical yield of 55 g and an actual yield of 33 g. Which substitution into the percent yield equation is correct?
A reaction has a theoretical yield of 32.0 g. The experiment collects 29.6 g of product. What is the percent yield? Give your answer as a percentage to 3 significant figures.
Answer: 92.5 % KEY
also accepted: 92.5%
wrongValueDiagnostics
[
{
"value": 108.1,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 0.925,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
},
{
"value": 2.4,
"misconception": "you subtracted the yields — percent yield compares them by division, not by difference",
"feedback": "This answer is wrong — you subtracted the yields — percent yield compares them by division, not by difference."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nactual yield = 29.6 g\ntheoretical yield = 32.0 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nSubstitute in the values, and calculate:\npercent yield = (29.6 ÷ 32.0) × 100\npercent yield = 92.5%"
}
STO-030 · medium · application · seq 9 · authored (parallel numeric slot; fresh taught-class precipitation SnCl₂ + 2KOH → Sn(OH)₂ + 2KCl with the measured solute carrying coefficient 2 — the spare p22 halving pattern on a fresh instance; KOH is the measured limiting reactant and the tin salt is in excess, so no amphoteric re-dissolution and no acid–base framing; values 0.44 L / 0.15 M ledger-free, exact product) · numeric
0.44 L of 0.15 M potassium hydroxide solution reacts with excess tin(II) chloride by SnCl₂ + 2KOH → Sn(OH)₂ + 2KCl. How many moles of Sn(OH)₂ form? Give your answer in moles to 2 significant figures.
Answer: 0.033 mol KEY
also accepted: 0.033 mol
wrongValueDiagnostics
[
{
"value": 0.066,
"misconception": "you stopped at the moles of KOH — the mole ratio of Sn(OH)₂ to KOH is 1 : 2",
"feedback": "This answer is wrong — you stopped at the moles of KOH — the mole ratio of Sn(OH)₂ to KOH is 1 : 2."
},
{
"value": 0.13,
"misconception": "you flipped the mole ratio and doubled the KOH moles instead of halving them",
"feedback": "This answer is wrong — you flipped the mole ratio and doubled the KOH moles instead of halving them."
}
]
feedback
{
"keyExplanation": "Step 1 — moles of dissolved reactant:\nn(KOH) = M × V\nn(KOH) = 0.15 × 0.44 = 0.066 mol\nStep 2 — moles of product with the mole ratio:\nmoles of Sn(OH)₂ = 0.066 × (1 ÷ 2)\nmoles of Sn(OH)₂ = 0.033 mol"
}
A student's synthesis has a theoretical yield of 165 g. The experiment collects 33 g of product. What is the percent yield? Give your answer as a percentage to 3 significant figures.
Answer: 20.0 % KEY
also accepted: 20.0%
wrongValueDiagnostics
[
{
"value": 500,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 0.2,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
},
{
"value": 132,
"misconception": "you subtracted the actual yield from the theoretical yield and reported the difference as a percent",
"feedback": "This answer is wrong — you subtracted the actual yield from the theoretical yield and reported the difference as a percent."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nactual yield = 33 g\ntheoretical yield = 165 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nSubstitute in the values, and calculate:\npercent yield = (33 ÷ 165) × 100\npercent yield = 20.0%\nThe actual yield is far below the theoretical yield, so the percentage must come out well below 100% — an answer like 500% fails that check."
}
STO-027 · medium · application · seq 11 · authored (fresh pair 325 g at 16% -> exact zero-free 52 g; '16' ≠ corpus '16.0'/'160'/'16000' string-forms; seed p12's structure and diagnostic classes kept; tail = whitelisted convention phrase) · numeric
A synthesis run has a theoretical yield of 325 g, and the method reliably gives a 16% yield. What mass of product should the run expect to collect? Give your answer in grams to 2 significant figures.
Answer: 52 g KEY
also accepted: 52 g
wrongValueDiagnostics
[
{
"value": 2031.25,
"misconception": "you divided the theoretical yield by the percent fraction instead of multiplying — a below-100% yield must give an actual yield below the theoretical 325 g",
"feedback": "This answer is wrong — you divided the theoretical yield by the percent fraction instead of multiplying — a below-100% yield must give an actual yield below the theoretical 325 g."
},
{
"value": 5200,
"misconception": "you forgot to divide the percent yield by 100 — thousands of grams from a 325 g prediction is the signal",
"feedback": "This answer is wrong — you forgot to divide the percent yield by 100 — thousands of grams from a 325 g prediction is the signal."
},
{
"value": 309,
"misconception": "you subtracted the percent number from the theoretical yield as if it were a mass in grams — 16 is a percentage, not a mass",
"feedback": "This answer is wrong — you subtracted the percent number from the theoretical yield as if it were a mass in grams — 16 is a percentage, not a mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\npercent yield = 16%\ntheoretical yield = 325 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nMake the actual yield the subject:\nactual yield = (percent yield ÷ 100) × theoretical yield\nSubstitute in the values, and calculate:\nactual yield = (16 ÷ 100) × 325\nactual yield = 52 g\nSanity check: 52 g is below the theoretical 325 g — a percent yield below 100% must give an actual yield below the theoretical yield."
}
STO-028 · medium · application · seq 12 · authored (fresh pair 97/108, both strings prescan-clean; over-cap 108 >= 103 so rounding never excuses it; plant two-shift surface; seed p15's structure, key position, and all three distractor classes kept; word-dressed cap distractors — no extra numerals) · mcq4
A plant runs one reaction on two shifts and logs a yield for each. The day shift logs 97%; the night shift logs 108%. Which log entry needs rechecking?
The night shift's — a genuine percent yield cannot rise above 100%. KEY
The day shift's — a 97% yield is too good to be genuine.
Both entries — real production runs never come anywhere near either value.
Neither entry — values several points above 100% are normal shift-to-shift variation.
STO-026 · medium · application · seq 13 · authored (fresh exact-division dataset pair 360/198 vs 240/174 -> 55% vs 72.5%, winner second-listed with fewer grams; all strings prescan-clean; re-valued from an initial 68% after finding '68' in row04; seed p11's structure and all three distractor classes kept; results-only options) · mcq4
A student compares two trials. Trial 1 has a theoretical yield of 360 g and collects 198 g of product. Trial 2 has a theoretical yield of 240 g and collects 174 g. Which trial has the higher percent yield?
Trial 2 — 72.5% compared with 55%. KEY
Trial 1 — more collected grams always mean a higher percent yield.
Trial 1 — 182% compared with 138%.
The two trials tie — any shortfall below the prediction counts equally, whatever its size.
STO-029 · medium · application · seq 14 · authored (fresh decomposition family MnCO₃ -> MnO — manganese off every burned list; seed p18's corpus-clean decomposition template; exact route 0.500 mol -> T 35.45 g -> 3/5 = 60.0%; all stem strings ledger-free; answer-side 60.0 echoes an STO-029 fluency value only — recorded for the blind round, stem carries no 60) · numeric
57.45 g of MnCO₃ is heated and decomposes by MnCO₃ → MnO + CO₂. 21.27 g of MnO is collected. What is the percent yield of MnO? (M of MnCO₃ = 114.9 g/mol; M of MnO = 70.9 g/mol.) Give your answer as a percentage to 3 significant figures.
Answer: 60.0 % KEY
also accepted: 60.0%
wrongValueDiagnostics
[
{
"value": 37.0,
"misconception": "you divided the product mass by the reactant mass — the comparison is with the theoretical yield, 35.45 g",
"feedback": "This answer is wrong — you divided the product mass by the reactant mass — the comparison is with the theoretical yield, 35.45 g."
},
{
"value": 167,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
}
]
feedback
{
"keyExplanation": "Step 1 — convert the reactant mass to moles:\nn(MnCO₃) = 57.45 / 114.9 = 0.500 mol\nStep 2 — convert to moles of product with the mole ratio:\nmoles of MnO = 0.500 × (1 ÷ 1) = 0.500 mol\nStep 3 — convert the product moles to mass (the theoretical yield):\ntheoretical yield = 0.500 × 70.9 = 35.45 g\nStep 4 — calculate the percent yield:\npercent yield = (21.27 ÷ 35.45) × 100\npercent yield = 60.0%"
}
STO-030 · hard · application · seq 15 · authored (fresh taught-class precipitation: lead(II) nitrate + ammonium sulfate → PbSO₄, 1:1 from the dissolved reactant; lead(II) nitrate is classroom-standard and outside the banned set per the blueprint, 0.85 M is comfortably inside its solubility; 'nitrate solution is mixed with' dodges WE1's 'nitrate solution reacts with' 4-gram; no material repeats within form B — row09-B is tin/KOH, row14-B is MnCO₃) · numeric
800 mL of 0.85 M lead(II) nitrate solution is mixed with excess ammonium sulfate. The reaction is Pb(NO₃)₂ + (NH₄)₂SO₄ → PbSO₄ + 2NH₄NO₃. How many moles of PbSO₄ form? Give your answer in moles to 2 significant figures.
Answer: 0.68 mol KEY
also accepted: 0.68 mol
wrongValueDiagnostics
[
{
"value": 680.0,
"misconception": "you used the volume in milliliters — n = M × V needs liters, and 800 mL is 0.800 L",
"feedback": "This answer is wrong — you used the volume in milliliters — n = M × V needs liters, and 800 mL is 0.800 L."
},
{
"value": 0.85,
"misconception": "you used the molarity as the mole amount without multiplying by the volume",
"feedback": "This answer is wrong — you used the molarity as the mole amount without multiplying by the volume."
},
{
"value": 0.94,
"misconception": "you divided the volume by the molarity instead of multiplying them",
"feedback": "This answer is wrong — you divided the volume by the molarity instead of multiplying them."
}
]
feedback
{
"keyExplanation": "Convert the volume to liters first: 800 mL = 0.800 L.\nStep 1 — moles of dissolved reactant:\nn(Pb(NO₃)₂) = M × V\nn(Pb(NO₃)₂) = 0.85 × 0.800 = 0.68 mol\nStep 2 — moles of product with the mole ratio:\nmoles of PbSO₄ = 0.68 × (1 ÷ 1)\nmoles of PbSO₄ = 0.68 mol\nA 680 mol answer fails the sanity check — less than a liter of solution holds well under a mole of dissolved salt at this molarity."
}
STO-029 · hard · application · seq 16 · authored (fresh zirconium combustion — flashbulb chemistry, no reactor framing; tempting pattern: collected 4.312 g just under the 4.56 g burned, naive comparison 94.6% vs true 70.0%; Zr 91.2 / ZrO₂ 123.2 real at 1 decimal and internally consistent; exact route 0.0500 mol -> 6.16 g -> 70.0%; masses ≤12 unscanned, molar-mass strings 91.2/123.2 ledger-free and disjoint from row14's titanium 47.9/79.9 which forced this redesign) · numeric
4.56 g of zirconium powder is burned in excess O₂ by Zr + O₂ → ZrO₂, and 4.312 g of ZrO₂ is collected. What is the percent yield? (M of Zr = 91.2 g/mol; M of ZrO₂ = 123.2 g/mol.) Give your answer as a percentage to 3 significant figures.
Answer: 70.0 % KEY
also accepted: 70.0%
wrongValueDiagnostics
[
{
"value": 142.9,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 94.6,
"misconception": "you compared the product mass with the reactant mass — the theoretical yield is 6.16 g of ZrO₂, not 4.56 g",
"feedback": "This answer is wrong — you compared the product mass with the reactant mass — the theoretical yield is 6.16 g of ZrO₂, not 4.56 g."
},
{
"value": 0.7,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
}
]
feedback
{
"keyExplanation": "Step 1 — convert the reactant mass to moles:\nn(Zr) = 4.56 / 91.2 = 0.0500 mol\nStep 2 — convert to moles of product with the mole ratio:\nmoles of ZrO₂ = 0.0500 × (1 ÷ 1) = 0.0500 mol\nStep 3 — convert the product moles to mass (the theoretical yield):\ntheoretical yield = 0.0500 × 123.2 = 6.16 g\nStep 4 — calculate the percent yield:\npercent yield = (4.312 ÷ 6.16) × 100\npercent yield = 70.0%\nThe collected ZrO₂ mass sitting just below the zirconium mass is a coincidence of masses — the oxide carries added oxygen mass, and the yield comparison is always against the theoretical yield."
}
STO-024 · hard · transfer · seq 17 · authored (matter-became-heat claim rotation per blueprint; exothermic-flask surface; four distinct accounts — conservation key + endorse + atoms-destroyed + double-counting; gap mass 3 g <= 12; instance-free) · mcq4
A strongly exothermic run warms its flask so much that the mixture stays hot for several minutes. The actual yield measures 3 g below the theoretical yield, and a lab partner claims the missing 3 g of matter was converted into the heat the reaction released. What is the correct account?
No matter became heat — the missing mass is matter that ended up elsewhere, such as unreacted starting material, side products, or residue left behind on equipment. KEY
The partner is right — an exothermic reaction turns a small fraction of its matter into the energy it gives off, so the hotter the run, the lighter the product.
The high temperature destroyed some of the atoms before they could form product.
The theoretical yield counted part of the starting amount twice, so the prediction came out too high.
STO-027 · hard · transfer · seq 18 · authored (fresh triple: theoretical 350 g at 34% -> expected 119 g vs 131 g promised, exact, all strings prescan-clean; complement 231 g coherent — above the promise since 34% < 50; supplier-promise surface; seed p14's structure, No verdict, key position, and all three distractor classes kept) · mcq4
A supplier promises a customer 131 g of a compound. The planned synthesis has a theoretical yield of 350 g, and the reaction typically gives a 34% yield. Is the synthesis expected to cover the promise?
No — the expected actual yield is 119 g, below the 131 g promised. KEY
Yes — with a 350 g theoretical yield the synthesis is certain to hand over the 131 g promised.
Yes — the expected actual yield is 231 g, well above the 131 g promised.
It cannot be judged — without the molar masses of the reactants and the compound, no expected mass can be found.
STO-028 · hard · transfer · seq 19 · authored (doubled-excess dressing per blueprint; seed p16's construct, key account, and all three wrong-account classes kept; numeral surface = the taught constant 100 only; instance-free) · mcq4
A classmate's plan reads: 'Double the excess reactant and the run will finally beat a 100% yield, because twice the extra reactant means extra product.' What is wrong with the plan?
The theoretical yield already counts all of the limiting reactant, so doubling the excess adds nothing to the possible product. KEY
Nothing is wrong — doubling the excess really does push careful runs past a 100% yield.
The plan fails only because doubling the excess makes the reaction run too fast to control safely in the lab.
The plan fails because doubling the excess cuts the theoretical yield instead.
STO-029 · hard · transfer · seq 20 · authored (fresh metal-oxidation pair per the blueprint's B–E guidance: molybdenum burns in excess O₂ to molybdenum(VI) oxide, 1:1 metal-to-oxide, both molar masses at 1 dp; all six stem values and the four option values verified spec- and inventory-clean; raw-grams mislead holds (29.52 > 28.44 while 41% < 79%); molybdenum is non-lexicon and appears nowhere else in the gate) · mcq4
Two runs of 2Mo + 3O₂ → 2MoO₃ burn different amounts of molybdenum in excess O₂ (M of Mo = 96.0 g/mol; M of MoO₃ = 144.0 g/mol). Run 1: 48.0 g of Mo, 29.52 g of MoO₃ collected. Run 2: 24.0 g of Mo, 28.44 g of MoO₃ collected. Which run has the higher percent yield?
Run 2 — 79% compared with 41%. KEY
Run 1 — collecting more grams of MoO₃ (29.52 g against 28.44 g) guarantees the higher percent yield.
Run 1 — 244% compared with 127%.
The runs tie — the same reaction always gives the same percent yield, whatever the scale of the run.
STO-030 · medium · application · seq 21 · authored (FRQ-1 form B: fresh taught-class precipitation Sr(NO₃)₂ + (NH₄)₂CO₃ → SrCO₃ + 2NH₄NO₃ with the DOUBLING ratio read off the printed coefficients; strontium(III) is classroom-standard and absent from every scope corpus; exact product 0.0476 mol → 0.048 mol at 2 s.f.) · frq
0.28 L of 0.17 M strontium nitrate solution is treated with excess ammonium carbonate by Sr(NO₃)₂ + (NH₄)₂CO₃ → SrCO₃ + 2NH₄NO₃.
0.28 L of 0.17 M strontium nitrate solution is treated with excess ammonium carbonate by Sr(NO₃)₂ + (NH₄)₂CO₃ → SrCO₃ + 2NH₄NO₃.
(a) Calculate how many moles of Sr(NO₃)₂ are dissolved in the 0.28 L of solution. Show your working.
(b) Use the mole ratio from the balanced equation to calculate how many moles of SrCO₃ form. Show your working. Give your answer in moles to 2 significant figures.
Part a. Calculate how many moles of Sr(NO₃)₂ are dissolved in the 0.28 L of solution. Show your working.
Model: Values: V = 0.28 L of solution, M = 0.17 mol/L — M is the molarity of the strontium nitrate solution, not a molar mass.
n = M × V
n = 0.17 × 0.28
n = 0.0476 mol of strontium nitrate dissolved.
Part b. Use the mole ratio from the balanced equation to calculate how many moles of SrCO₃ form. Show your working. Give your answer in moles to 2 significant figures.
Model: From the balanced equation, 1 mol of strontium carbonate forms for every 1 mol of strontium nitrate.
moles of strontium carbonate = 0.0476 × (1 ÷ 1)
moles of strontium carbonate = 0.0476 mol
To 2 significant figures: 0.048 mol of strontium carbonate. The product stays in moles.
Full model answer. (a) Values: V = 0.28 L of solution, M = 0.17 mol/L — M is the molarity of the strontium nitrate solution, not a molar mass.
n = M × V
n = 0.17 × 0.28
n = 0.0476 mol of strontium nitrate dissolved.
(b) From the balanced equation, 1 mol of strontium carbonate forms for every 1 mol of strontium nitrate.
moles of strontium carbonate = 0.0476 × (1 ÷ 1)
moles of strontium carbonate = 0.0476 mol
To 2 significant figures: 0.048 mol of strontium carbonate. The product stays in moles.
STO-029 · hard · application · seq 22 · authored (FRQ-2 form B: bismuth burn, 4:2 ratio forces the shown ratio step; UNSEEDED) · frq
A crucible holding 62.7 g of bismuth powder is ignited; the bismuth burns in excess O₂ by 4Bi + 3O₂ → 2Bi₂O₃. After cooling, 27.96 g of Bi₂O₃ is collected.
A crucible holding 62.7 g of bismuth powder is ignited; the bismuth burns in excess O₂ by 4Bi + 3O₂ → 2Bi₂O₃. After cooling, 27.96 g of Bi₂O₃ is collected.
(a) Calculate the theoretical yield of Bi₂O₃. Show every step of your working. Report the theoretical yield to 3 significant figures.
(b) Calculate the percent yield of Bi₂O₃. Give your answer as a percentage to 3 significant figures.
Molar masses (exact): Bi 209.0 g/mol, Bi₂O₃ 466.0 g/mol.
Write your answers in plain numbers and words; no special symbols are needed.
Part a. Calculate the theoretical yield of Bi₂O₃. Show every step of your working. Report the theoretical yield to 3 significant figures.
Model: Values: mass of bismuth = 62.7 g; molar mass of bismuth = 209.0 g/mol; molar mass of bismuth oxide = 466.0 g/mol.
Equation: n = m ÷ M
Substitute: n = 62.7 ÷ 209.0
Calculate: n = 0.300 mol of bismuth
Mole ratio from the balanced equation: 2 mol of bismuth oxide forms from 4 mol of bismuth, so n = 0.300 × (2 ÷ 4) = 0.150 mol of bismuth oxide
Equation: m = n × M
Substitute: m = 0.150 × 466.0
Calculate: theoretical yield = 69.90 g, reported as 69.9 g to 3 significant figures
Part b. Calculate the percent yield of Bi₂O₃. Give your answer as a percentage to 3 significant figures.
Molar masses (exact): Bi 209.0 g/mol, Bi₂O₃ 466.0 g/mol.
Write your answers in plain numbers and words; no special symbols are needed.
Model: Values: actual yield = 27.96 g; theoretical yield = 69.90 g.
Equation: percent yield = (actual yield ÷ theoretical yield) × 100
Substitute: percent yield = (27.96 ÷ 69.90) × 100
Calculate: percent yield = 40.0% (the rounded 69.9 g gives the same 40.0%)
Full model answer. (a) Values: mass of bismuth = 62.7 g; molar mass of bismuth = 209.0 g/mol; molar mass of bismuth oxide = 466.0 g/mol.
Equation: n = m ÷ M
Substitute: n = 62.7 ÷ 209.0
Calculate: n = 0.300 mol of bismuth
Mole ratio from the balanced equation: 2 mol of bismuth oxide forms from 4 mol of bismuth, so n = 0.300 × (2 ÷ 4) = 0.150 mol of bismuth oxide
Equation: m = n × M
Substitute: m = 0.150 × 466.0
Calculate: theoretical yield = 69.90 g, reported as 69.9 g to 3 significant figures
(b) Values: actual yield = 27.96 g; theoretical yield = 69.90 g.
Equation: percent yield = (actual yield ÷ theoretical yield) × 100
Substitute: percent yield = (27.96 ÷ 69.90) × 100
Calculate: percent yield = 40.0% (the rounded 69.9 g gives the same 40.0%)
While planning tomorrow's experiment, a technician works through a stoichiometry calculation and finds that at most 4.9 g of product can form. Which label does the 4.9 g carry?
The theoretical yield. KEY
The actual yield.
Both yields at once.
Neither — a number worked out before any product exists cannot be a yield.
STO-025 · easy · recall · seq 2 · authored (recall probe C: which two quantities enter; 'quantities' chosen over the WE's 'amounts' to keep the stem gram-distinct from WE2) · mcq4
A student has finished an experiment and wants to report the percent yield. Which two quantities must the student have in hand?
The actual yield and the theoretical yield. KEY
The actual yield and the mass of reactant supplied.
The theoretical yield and the mass of reactant left over.
The actual yield and the balanced chemical equation.
A group's report on a single synthesis states: 'our calculation gave 57 g; the mass of product we obtained and weighed was 49 g.' Which value is the group's actual yield?
The 49 g, because it is the amount that was actually measured. KEY
The 57 g, because the larger value shows what the reaction can really do.
Both values, because the report lists them for the same product.
Neither value — the actual yield is the 8 g the group did not collect.
Two revision cards show different versions of the percent yield equation. Card 1 reads percent yield = (actual yield ÷ theoretical yield) × 100. Card 2 reads percent yield = (actual yield ÷ theoretical yield) ÷ 100. Which card is right?
Card 1 — the fraction is multiplied by 100 to turn it into a percentage. KEY
Card 2 — a percentage means dividing by 100, so the fraction is divided by 100.
Both — multiplying or dividing the fraction by 100 gives the same percentage.
Neither — the fraction is already a percentage and needs no factor of 100.
STO-024 · medium · application · seq 6 · authored (incomplete-reaction rotation: stopped-early stirred-flask surface per blueprint; wording gram-distinct from WE2's stopped-reaction lines; instance-light — no named substances) | blind-wave1 2026-09-08: evidence-scoped ask (F2 class sweep) · mcq4
Two solutions react slowly in a stirred flask. The class period ends, so a student stops the run and collects the product even though a quick test shows that much of both starting solutions has not yet reacted. The actual yield finishes below the theoretical yield. Which cause does this observation directly show?
The reaction stopped before completion. KEY
A side reaction consumed reactant.
Product was lost on transfer.
Atoms were destroyed while the reaction ran.
STO-025 · medium · application · seq 7 · authored (fresh exact-division yield pair 88/66, ratio 3:4, both strings prescan-clean; lab-report surface; seed p6's structure and distractor classes kept) · mcq4
A student's lab report lists a theoretical yield of 88 g and an actual yield of 66 g. Which substitution into the percent yield equation is correct?
A reaction in a research lab has a theoretical yield of 24.8 g. The lab group collects 21.7 g of product. What is the percent yield? Give your answer as a percentage to 3 significant figures.
Answer: 87.5 % KEY
also accepted: 87.5%
wrongValueDiagnostics
[
{
"value": 114.3,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 0.875,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
},
{
"value": 3.1,
"misconception": "you subtracted the yields — percent yield compares them by division, not by difference",
"feedback": "This answer is wrong — you subtracted the yields — percent yield compares them by division, not by difference."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nactual yield = 21.7 g\ntheoretical yield = 24.8 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nSubstitute in the values, and calculate:\npercent yield = (21.7 ÷ 24.8) × 100\npercent yield = 87.5%"
}
STO-030 · medium · application · seq 9 · authored (parallel numeric slot; fresh taught-class precipitation MnCl₂ + K₂CO₃ → MnCO₃ + 2KCl, 1 : 1 asked ratio with the 2KCl coefficient as the doubling lure — the seed p21 construct on a fresh instance; values 0.55 L / 0.14 M ledger-free, exact product) · numeric
0.55 L of 0.14 M manganese(II) chloride solution reacts with excess potassium carbonate by MnCl₂ + K₂CO₃ → MnCO₃ + 2KCl. How many moles of MnCO₃ form? Give your answer in moles to 2 significant figures.
Answer: 0.077 mol KEY
also accepted: 0.077 mol
wrongValueDiagnostics
[
{
"value": 0.14,
"misconception": "you used the molarity as the mole amount without multiplying by the volume",
"feedback": "This answer is wrong — you used the molarity as the mole amount without multiplying by the volume."
},
{
"value": 0.15,
"misconception": "you doubled the moles with KCl's coefficient — MnCO₃'s ratio to MnCl₂ is 1 : 1",
"feedback": "This answer is wrong — you doubled the moles with KCl's coefficient — MnCO₃'s ratio to MnCl₂ is 1 : 1."
}
]
feedback
{
"keyExplanation": "Step 1 — moles of dissolved reactant:\nn(MnCl₂) = M × V\nn(MnCl₂) = 0.14 × 0.55 = 0.077 mol\nStep 2 — moles of product with the mole ratio:\nmoles of MnCO₃ = 0.077 × (1 ÷ 1)\nmoles of MnCO₃ = 0.077 mol"
}
STO-026 · medium · application · seq 10 · authored (fresh low-yield pair 41.6 g / 128 g -> 32.5%; strings '41.6'/'128'/'32.5' ledger-free ('41.6' != corpus '16.0'); exact division 41.6/128 = 0.325; product-category surface, no material words) · numeric
A pigment preparation has a theoretical yield of 128 g. The experiment collects 41.6 g of product. What is the percent yield? Give your answer as a percentage to 3 significant figures.
Answer: 32.5 % KEY
also accepted: 32.5%
wrongValueDiagnostics
[
{
"value": 308,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 0.325,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
},
{
"value": 86.4,
"misconception": "you subtracted the actual yield from the theoretical yield and reported the difference as a percent",
"feedback": "This answer is wrong — you subtracted the actual yield from the theoretical yield and reported the difference as a percent."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nactual yield = 41.6 g\ntheoretical yield = 128 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nSubstitute in the values, and calculate:\npercent yield = (41.6 ÷ 128) × 100\npercent yield = 32.5%\nThe actual yield is far below the theoretical yield, so the percentage must come out well below 100% — an answer like 308% fails that check."
}
STO-027 · medium · application · seq 11 · authored (fresh pair 300 g at 17% -> exact zero-free 51 g; '17' ≠ corpus '17.0' string-form; seed p12's structure and diagnostic classes kept; tail = whitelisted convention phrase) · numeric
A manufacturing run has a theoretical yield of 300 g, and the process reliably delivers a 17% yield. What mass of product should the run expect to collect? Give your answer in grams to 2 significant figures.
Answer: 51 g KEY
also accepted: 51 g
wrongValueDiagnostics
[
{
"value": 1764.7,
"misconception": "you divided the theoretical yield by the percent fraction instead of multiplying — a below-100% yield must give an actual yield below the theoretical 300 g",
"feedback": "This answer is wrong — you divided the theoretical yield by the percent fraction instead of multiplying — a below-100% yield must give an actual yield below the theoretical 300 g."
},
{
"value": 5100,
"misconception": "you forgot to divide the percent yield by 100 — thousands of grams from a 300 g prediction is the signal",
"feedback": "This answer is wrong — you forgot to divide the percent yield by 100 — thousands of grams from a 300 g prediction is the signal."
},
{
"value": 283,
"misconception": "you subtracted the percent number from the theoretical yield as if it were a mass in grams — 17 is a percentage, not a mass",
"feedback": "This answer is wrong — you subtracted the percent number from the theoretical yield as if it were a mass in grams — 17 is a percentage, not a mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\npercent yield = 17%\ntheoretical yield = 300 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nMake the actual yield the subject:\nactual yield = (percent yield ÷ 100) × theoretical yield\nSubstitute in the values, and calculate:\nactual yield = (17 ÷ 100) × 300\nactual yield = 51 g\nSanity check: 51 g is below the theoretical 300 g — a percent yield below 100% must give an actual yield below the theoretical yield."
}
STO-028 · medium · application · seq 12 · authored (fresh pair 91/116, both strings prescan-clean; over-cap 116 >= 103; teaching-lab two-bench surface; seed p15's structure and distractor classes kept; word-dressed cap distractors) · mcq4
Two benches in a teaching lab ran identical syntheses. Bench 1's report gives a 91% yield; Bench 2's report gives a 116% yield. Which report needs rechecking?
STO-026 · medium · application · seq 13 · authored (fresh exact-division dataset pair 520/195 vs 280/189 -> 37.5% vs 67.5%, winner first-listed with fewer grams; plant-batch surface; loser re-valued from an initial 38% after finding '38' in row04; all strings prescan-clean; seed p11's structure and distractor classes kept; results-only options) · mcq4
A production plant compares two batches. Batch 1 has a theoretical yield of 520 g and collects 195 g of product. Batch 2 has a theoretical yield of 280 g and collects 189 g. Which batch has the higher percent yield?
Batch 2 — 67.5% compared with 37.5%. KEY
Batch 1 — the batch that collects more grams always has the higher percent yield.
Batch 1 — 267% compared with 148%.
The two batches tie — being below prediction is all that matters, not by how much.
STO-029 · medium · application · seq 14 · authored (fresh combustion family Ti -> TiO₂; spec-mandated excess convention 'burns in excess O₂ by' = gate_config idiom A-U8G3-2; exact route 0.800 mol -> T 63.92 g -> 5/8 = 62.5%; collected > reactant mass is inherent oxide growth, not a seq-16-style near-equality; all stem strings ledger-free) · numeric
38.32 g of titanium powder burns in excess O₂ by Ti + O₂ → TiO₂, and 39.95 g of TiO₂ is collected. What is the percent yield? (M of Ti = 47.9 g/mol; M of TiO₂ = 79.9 g/mol.) Give your answer as a percentage to 3 significant figures.
Answer: 62.5 % KEY
also accepted: 62.5%
wrongValueDiagnostics
[
{
"value": 104,
"misconception": "you divided the product mass by the reactant mass — the comparison is with the theoretical yield, 63.92 g; comparing with the metal's mass even pushes the result past 100, which a genuine yield can never do",
"feedback": "This answer is wrong — you divided the product mass by the reactant mass — the comparison is with the theoretical yield, 63.92 g; comparing with the metal's mass even pushes the result past 100, which a genuine yield can never do."
},
{
"value": 160,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
}
]
feedback
{
"keyExplanation": "Step 1 — convert the reactant mass to moles:\nn(Ti) = 38.32 / 47.9 = 0.800 mol\nStep 2 — convert to moles of product with the mole ratio:\nmoles of TiO₂ = 0.800 × (1 ÷ 1) = 0.800 mol\nStep 3 — convert the product moles to mass (the theoretical yield):\ntheoretical yield = 0.800 × 79.9 = 63.92 g\nStep 4 — calculate the percent yield:\npercent yield = (39.95 ÷ 63.92) × 100\npercent yield = 62.5%"
}
STO-030 · hard · application · seq 15 · authored (2026-09-08 crash-resume rebuild: the shipped form C used cobalt(II) chloride at 0.15 M / 460 mL — 'cobalt' collided cross-form with E#9 and within-form with the C#21 FRQ, and '0.15' with B#9's molarity. Rebuilt as barium chloride + ammonium carbonate → BaCO₃ (classic classroom precipitation, dilute 0.13 M reagent); 'barium' is non-lexicon and the only other barium item is A#9, a different form; the stem's 'holds ... solution at a concentration of' phrasing deliberately avoids A#9's 'M barium chloride solution reacts' 4-gram windows; 700 mL / 0.13 M chosen spec-clean and inventory-clean ('380'/'0.50' candidates sat in spec text); mL trap, 1:1 ratio, three diagnostic classes, scaffold and sanity line kept; answer 0.091 mol exact at 2 sig figs) · numeric
A beaker holds 700 mL of barium chloride solution at a concentration of 0.13 M. Excess ammonium carbonate is added, and the mixture reacts by BaCl₂ + (NH₄)₂CO₃ → BaCO₃ + 2NH₄Cl. How many moles of BaCO₃ form? Give your answer in moles to 2 significant figures.
Answer: 0.091 mol KEY
also accepted: 0.091 mol
wrongValueDiagnostics
[
{
"value": 91.0,
"misconception": "you used the volume in milliliters — n = M × V needs liters, and 700 mL is 0.700 L",
"feedback": "This answer is wrong — you used the volume in milliliters — n = M × V needs liters, and 700 mL is 0.700 L."
},
{
"value": 0.13,
"misconception": "you used the molarity as the mole amount without multiplying by the volume",
"feedback": "This answer is wrong — you used the molarity as the mole amount without multiplying by the volume."
},
{
"value": 5.38,
"misconception": "you divided the volume by the molarity instead of multiplying them",
"feedback": "This answer is wrong — you divided the volume by the molarity instead of multiplying them."
}
]
feedback
{
"keyExplanation": "Convert the volume to liters first: 700 mL = 0.700 L.\nStep 1 — moles of dissolved reactant:\nn(BaCl₂) = M × V\nn(BaCl₂) = 0.13 × 0.700 = 0.091 mol\nStep 2 — moles of product with the mole ratio:\nmoles of BaCO₃ = 0.091 × (1 ÷ 1)\nmoles of BaCO₃ = 0.091 mol\nA 91 mol answer fails the sanity check — under a liter of a dilute solution holds well under a mole of dissolved salt."
}
STO-029 · hard · application · seq 16 · authored (fresh silicon combustion; tempting pattern: collected 5.409 g just under the 5.62 g of silicon burned, naive comparison 96.2% vs true 45.0%; Si 28.1 / SiO₂ 60.1 real at 1 decimal and internally consistent; exact route 0.200 mol -> 12.02 g -> 45.0%; masses ≤12 unscanned, molar-mass strings ledger-free — '28.1' ≠ corpus '28.0', '60.1' ≠ corpus '60') · numeric
5.62 g of silicon burns in excess O₂ by Si + O₂ → SiO₂, and 5.409 g of SiO₂ is collected. What is the percent yield? (M of Si = 28.1 g/mol; M of SiO₂ = 60.1 g/mol.) Give your answer as a percentage to 3 significant figures.
Answer: 45.0 % KEY
also accepted: 45.0%
wrongValueDiagnostics
[
{
"value": 222.2,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 96.2,
"misconception": "you compared the product mass with the reactant mass — the theoretical yield is 12.02 g of SiO₂, not 5.62 g",
"feedback": "This answer is wrong — you compared the product mass with the reactant mass — the theoretical yield is 12.02 g of SiO₂, not 5.62 g."
},
{
"value": 0.45,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
}
]
feedback
{
"keyExplanation": "Step 1 — convert the reactant mass to moles:\nn(Si) = 5.62 / 28.1 = 0.200 mol\nStep 2 — convert to moles of product with the mole ratio:\nmoles of SiO₂ = 0.200 × (1 ÷ 1) = 0.200 mol\nStep 3 — convert the product moles to mass (the theoretical yield):\ntheoretical yield = 0.200 × 60.1 = 12.02 g\nStep 4 — calculate the percent yield:\npercent yield = (5.409 ÷ 12.02) × 100\npercent yield = 45.0%\nThe collected SiO₂ mass sitting just below the silicon mass is a coincidence of masses — the yield comparison is always against the theoretical yield, not the reactant mass."
}
STO-024 · hard · transfer · seq 17 · authored (prediction-counted-twice claim rotation per blueprint; group-record surface; four distinct accounts — counts-once conservation key + endorse + atoms-destroyed + matter-to-heat; gap mass 6 g <= 12; instance-free) · mcq4
A group's actual yield is 6 g short of the theoretical yield. One group member argues that the calculation must have counted some of the atoms twice, so the theoretical yield came out too high. What is the correct account?
The calculation counts each atom once — the missing 6 g is matter that ended up elsewhere, such as unreacted material, side products, or residue on equipment. KEY
The group member is right — theoretical yield calculations routinely double-count atoms, so every prediction comes out higher than a run could ever reach.
Atoms were destroyed while the reaction ran, so less mass was left to weigh.
The missing mass left the flask as the heat the reaction gave off.
STO-027 · hard · transfer · seq 18 · authored (VERDICT ROTATED to deliver: 450 g at 74% -> expected 333 g vs 310 g agreed, exact, all strings prescan-clean; percent > 50 puts the complement 117 g BELOW the agreement so the flipped-percentage option carries the wrong No verdict; compared-to-theoretical class dressed as the full-collection claim — same misconception, discriminable from the key; lab-agreement surface; two deliver-verdict options mirror the seed's two-Yes asymmetry, recorded for the blind round) · mcq4
A lab agrees to prepare 310 g of a compound for another class's investigation. The planned run has a theoretical yield of 450 g, and the reaction typically gives a 74% yield. Is the run expected to prepare enough to meet the agreement?
Yes — the expected actual yield is 333 g, above the 310 g agreed. KEY
Yes — the run will hand over the full 450 g planned, far more than the agreement needs.
No — the expected actual yield is only 117 g, far short of the 310 g agreed.
It cannot be judged — no expected mass can be calculated unless the molar masses of the compound and reactants are given.
STO-028 · hard · transfer · seq 19 · authored (dripped-in-excess dressing per blueprint; seed p16's construct, key account, and all three wrong-account classes kept; the speed distractor's 'fails only because' framing keeps it cleanly wrong whatever the drip rate does; numeral surface = 100 only; instance-free) · mcq4
A lab partner suggests: 'Keep dripping in more of the excess reactant during the whole run — the steady supply will carry the percent yield above 100%.' What is wrong with the suggestion?
The theoretical yield already counts all of the limiting reactant — however the excess arrives, product formation stops when the limiting reactant runs out. KEY
Nothing is wrong — a steady feed of excess reactant genuinely carries yields above 100%.
The suggestion fails only because adding the excess drop by drop makes the reaction too slow to finish.
The suggestion fails because each added portion of excess shrinks the theoretical yield a little further, so the percent can only fall.
STO-029 · hard · transfer · seq 20 · authored (fresh decomposition run per the blueprint's B–E guidance: strontium carbonate decomposes on strong heating to the oxide, 1:1, no second reactant so no excess statement is required; a low percent yield reads naturally as incomplete decomposition; 'strontium' is non-lexicon — its only other uses are A#15 and B#21, different forms; all values spec- and inventory-clean; raw-grams mislead holds (17.871 > 13.986 while 23% < 54%)) · mcq4
Two samples of strontium carbonate are heated strongly until they decompose by SrCO₃ → SrO + CO₂ (M of SrCO₃ = 147.6 g/mol; M of SrO = 103.6 g/mol). Run 1: 110.7 g of SrCO₃, 17.871 g of SrO collected. Run 2: 36.9 g of SrCO₃, 13.986 g of SrO collected. Which run has the higher percent yield?
Run 2 — 54% compared with 23%. KEY
Run 1 — collecting more grams of SrO (17.871 g against 13.986 g) means it must have the higher percent yield.
Run 1 — 435% compared with 185%.
The runs tie — a decomposition reaction fixes its percent yield, whatever the sample size.
STO-030 · medium · application · seq 21 · authored (FRQ-1 form C: fresh taught-class precipitation 2KOH + Mn(NO₃)₂ → Mn(OH)₂ + 2KNO₃ with the measured solute carrying coefficient 2 — the halving direction; KOH is the measured limiting reactant and the manganese salt is in excess, so no excess hydroxide exists; exact product 0.0713 mol → 0.071 mol at 2 s.f.) · frq
0.62 L of 0.23 M aqueous potassium hydroxide is treated with excess manganese(II) nitrate by 2KOH + Mn(NO₃)₂ → Mn(OH)₂ + 2KNO₃.
0.62 L of 0.23 M aqueous potassium hydroxide is treated with excess manganese(II) nitrate by 2KOH + Mn(NO₃)₂ → Mn(OH)₂ + 2KNO₃.
(a) Calculate how many moles of KOH are dissolved in the 0.62 L of solution. Show your working.
(b) Use the mole ratio from the balanced equation to calculate how many moles of Mn(OH)₂ form. Show your working. Give your answer in moles to 2 significant figures.
Part a. Calculate how many moles of KOH are dissolved in the 0.62 L of solution. Show your working.
Model: Values: V = 0.62 L of solution, M = 0.23 mol/L — M is the molarity of the potassium hydroxide solution, not a molar mass.
n = M × V
n = 0.23 × 0.62
n = 0.1426 mol of potassium hydroxide dissolved.
Part b. Use the mole ratio from the balanced equation to calculate how many moles of Mn(OH)₂ form. Show your working. Give your answer in moles to 2 significant figures.
Model: From the balanced equation, 1 mol of manganese(II) hydroxide forms for every 2 mol of potassium hydroxide.
moles of manganese(II) hydroxide = 0.1426 × (1 ÷ 2)
moles of manganese(II) hydroxide = 0.0713 mol
To 2 significant figures: 0.071 mol of manganese(II) hydroxide. The product stays in moles.
Full model answer. (a) Values: V = 0.62 L of solution, M = 0.23 mol/L — M is the molarity of the potassium hydroxide solution, not a molar mass.
n = M × V
n = 0.23 × 0.62
n = 0.1426 mol of potassium hydroxide dissolved.
(b) From the balanced equation, 1 mol of manganese(II) hydroxide forms for every 2 mol of potassium hydroxide.
moles of manganese(II) hydroxide = 0.1426 × (1 ÷ 2)
moles of manganese(II) hydroxide = 0.0713 mol
To 2 significant figures: 0.071 mol of manganese(II) hydroxide. The product stays in moles.
STO-029 · hard · application · seq 22 · authored (FRQ-2 form C: cerium burn, 1:1 route; UNSEEDED) · frq
A student ignites 70.05 g of cerium, which burns in excess O₂ by Ce + O₂ → CeO₂. After cooling, 51.63 g of CeO₂ is collected.
A student ignites 70.05 g of cerium, which burns in excess O₂ by Ce + O₂ → CeO₂. After cooling, 51.63 g of CeO₂ is collected.
(a) Calculate the theoretical yield of CeO₂. Show every step of your working. Report the theoretical yield to 3 significant figures.
(b) Calculate the percent yield of CeO₂. Give your answer as a percentage to 3 significant figures.
Molar masses (exact): Ce 140.1 g/mol, CeO₂ 172.1 g/mol.
Write your answers in plain numbers and words; no special symbols are needed.
Part a. Calculate the theoretical yield of CeO₂. Show every step of your working. Report the theoretical yield to 3 significant figures.
Model: Values: mass of cerium = 70.05 g; molar mass of cerium = 140.1 g/mol; molar mass of cerium dioxide = 172.1 g/mol.
Equation: n = m ÷ M
Substitute: n = 70.05 ÷ 140.1
Calculate: n = 0.5000 mol of cerium
Mole ratio from the balanced equation: 1 mol of cerium dioxide forms from 1 mol of cerium, so n = 0.5000 × (1 ÷ 1) = 0.5000 mol of cerium dioxide
Equation: m = n × M
Substitute: m = 0.5000 × 172.1
Calculate: theoretical yield = 86.05 g, reported as 86.1 g to 3 significant figures
Part b. Calculate the percent yield of CeO₂. Give your answer as a percentage to 3 significant figures.
Molar masses (exact): Ce 140.1 g/mol, CeO₂ 172.1 g/mol.
Write your answers in plain numbers and words; no special symbols are needed.
Model: Values: actual yield = 51.63 g; theoretical yield = 86.05 g.
Equation: percent yield = (actual yield ÷ theoretical yield) × 100
Substitute: percent yield = (51.63 ÷ 86.05) × 100
Calculate: percent yield = 60.0% (the rounded 86.1 g gives the same 60.0%)
Full model answer. (a) Values: mass of cerium = 70.05 g; molar mass of cerium = 140.1 g/mol; molar mass of cerium dioxide = 172.1 g/mol.
Equation: n = m ÷ M
Substitute: n = 70.05 ÷ 140.1
Calculate: n = 0.5000 mol of cerium
Mole ratio from the balanced equation: 1 mol of cerium dioxide forms from 1 mol of cerium, so n = 0.5000 × (1 ÷ 1) = 0.5000 mol of cerium dioxide
Equation: m = n × M
Substitute: m = 0.5000 × 172.1
Calculate: theoretical yield = 86.05 g, reported as 86.1 g to 3 significant figures
(b) Values: actual yield = 51.63 g; theoretical yield = 86.05 g.
Equation: percent yield = (actual yield ÷ theoretical yield) × 100
Substitute: percent yield = (51.63 ÷ 86.05) × 100
Calculate: percent yield = 60.0% (the rounded 86.1 g gives the same 60.0%)
A lab group finishes a reaction and records the dried product's mass: 10.6 g. Which yield label fits the 10.6 g?
The actual yield. KEY
The theoretical yield.
Both yields at once.
Neither — a yield must come from a calculation, and this number was weighed.
STO-025 · easy · recall · seq 2 · authored (recall probe D: which quantity is the yardstick; 'yardstick' vocabulary comes from the gateItem feedback register, which is corpus-excluded) · mcq4
In the percent yield equation, one of the two yields serves as the yardstick — the amount the other is measured against. Which one is the yardstick?
The theoretical yield — the measured amount is judged against the prediction. KEY
The actual yield — the predicted amount is judged against the measurement.
Whichever yield is larger — the bigger amount always serves as the base.
The mass of reactant supplied — both yields are judged against the starting material.
A group's percent yield lands in the high nineties — close to, but below, 100%. Could the value be genuine?
Yes — any value at or below 100% may be genuine, and a careful run can land close to the cap. KEY
No — real runs always lose enough product to keep genuine yields far below the nineties.
No — a value that close to the cap is itself the signal that the product was weighed while still wet.
Yes, but only because the 100% cap includes a built-in allowance for measurement error.
STO-023 · easy · application · seq 4 · authored (surface: results table described in prose — figure stays none by blueprint design; probe: theoretical yield; trust class inverted; values 68/61 ledger-verified) · mcq4
A results table for one reaction is described here in words: the first row lists 'mass of product predicted: 68 g'; the second row lists 'mass of product recovered and weighed: 61 g'. Which row holds the theoretical yield?
The 68 g row, because a predicted mass comes from a calculation, not from weighing. KEY
The 61 g row, because a directly weighed mass is more dependable than a prediction.
Both rows, because each mass belongs to the same product.
Neither row — the theoretical yield is the 7 g separating the two masses.
STO-025 · easy · recall · seq 5 · authored (defect rotation: subtraction instead of division; quiz surface; percent-loss lure as the Neither option; 100%-meaning feedback cites rule line 6 only, not STO-028's cap judgment; '100' -> A-U8G3-6) · mcq4
On a quiz, two classmates write the percent yield equation. Classmate 1 writes percent yield = (theoretical yield − actual yield) × 100. Classmate 2 writes percent yield = (actual yield ÷ theoretical yield) × 100. Who is right?
Classmate 2 — the actual yield is divided by the theoretical yield, then multiplied by 100. KEY
Classmate 1 — the gap between the predicted and measured amounts is what the percent reports.
Both — the two versions agree for a perfect experiment.
Neither — the difference between the yields should be divided by the theoretical yield.
STO-024 · medium · application · seq 6 · authored (side-reaction rotation: competing tarnish surface per blueprint; copper: cross-lesson WARN vs STO-030 (recorded, crash-resume wave-1) — no other material WARNs; copper(II) oxide is the genuine air-oxidation product and sulfide tarnish from trace burner-gas fumes is real chemistry) | blind-wave1 2026-09-08: evidence-scoped ask (F2 class sweep) · mcq4
A strip of copper is heated in air to convert it into copper(II) oxide. Reactive fumes leaking from a nearby burner reach the strip and react with part of the copper, converting that copper into a dark compound that is not copper(II) oxide. The dark material is separated from the oxide before it is weighed. The actual yield of copper(II) oxide ends up below the theoretical yield. Which cause does this observation directly show?
A side reaction consumed reactant. KEY
Product was lost on transfer.
The reaction stopped before completion.
Atoms of the copper were destroyed by the burner's heat.
STO-025 · medium · application · seq 7 · authored (fresh exact-division yield pair 32/12, ratio 3:8 — implied 37.5% is the row's non-integer-percent variant, never shown; '32' prescan-clean as a string ('32.1' is a different string), 12 in the scan-ignored 0..12 band; seed p6's structure and distractor classes kept) · mcq4
A run's theoretical yield is 32 g, and its actual yield is 12 g. Which substitution into the percent yield equation is correct?
A student's reaction has a theoretical yield of 52.5 g. The student collects 46.2 g of product. What is the percent yield? Give your answer as a percentage to 3 significant figures.
Answer: 88.0 % KEY
also accepted: 88.0%
wrongValueDiagnostics
[
{
"value": 113.6,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 0.88,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
},
{
"value": 6.3,
"misconception": "you subtracted the yields — percent yield compares them by division, not by difference",
"feedback": "This answer is wrong — you subtracted the yields — percent yield compares them by division, not by difference."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nactual yield = 46.2 g\ntheoretical yield = 52.5 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nSubstitute in the values, and calculate:\npercent yield = (46.2 ÷ 52.5) × 100\npercent yield = 88.0%"
}
STO-030 · medium · application · seq 9 · authored (parallel numeric slot; fresh taught-class precipitation 3NiCl₂ + 2K₃PO₄ → Ni₃(PO₄)₂ + 6KCl with the measured solute carrying coefficient 2 relative to the asked precipitate — the halving pattern on a second fresh instance; values 0.12 L / 0.90 M ledger-free, exact product) · numeric
0.12 L of 0.90 M potassium phosphate solution reacts with excess nickel(II) chloride by 3NiCl₂ + 2K₃PO₄ → Ni₃(PO₄)₂ + 6KCl. How many moles of Ni₃(PO₄)₂ form? Give your answer in moles to 2 significant figures.
Answer: 0.054 mol KEY
also accepted: 0.054 mol
wrongValueDiagnostics
[
{
"value": 0.11,
"misconception": "you stopped at the moles of K₃PO₄ — the mole ratio of Ni₃(PO₄)₂ to K₃PO₄ is 1 : 2",
"feedback": "This answer is wrong — you stopped at the moles of K₃PO₄ — the mole ratio of Ni₃(PO₄)₂ to K₃PO₄ is 1 : 2."
},
{
"value": 0.22,
"misconception": "you flipped the mole ratio and doubled the K₃PO₄ moles instead of halving them",
"feedback": "This answer is wrong — you flipped the mole ratio and doubled the K₃PO₄ moles instead of halving them."
}
]
feedback
{
"keyExplanation": "Step 1 — moles of dissolved reactant:\nn(K₃PO₄) = M × V\nn(K₃PO₄) = 0.90 × 0.12 = 0.108 mol\nStep 2 — moles of product with the mole ratio:\nmoles of Ni₃(PO₄)₂ = 0.108 × (1 ÷ 2)\nmoles of Ni₃(PO₄)₂ = 0.054 mol"
}
STO-026 · medium · application · seq 10 · authored (fresh low-yield pair 13.3 g / 76.0 g -> 17.5%, the register's low end; strings '13.3'/'76'/'17.5' ledger-free ('13.3' != corpus '133.3'; '17.5' != '17.0'); exact division 13.3/76 = 0.175; weighing surface replaces the collect-template) · numeric
An overnight reaction run has a theoretical yield of 76.0 g. The technician weighs the dried product: 13.3 g. What is the percent yield? Give your answer as a percentage to 3 significant figures.
Answer: 17.5 % KEY
also accepted: 17.5%
wrongValueDiagnostics
[
{
"value": 571,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 0.175,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
},
{
"value": 62.7,
"misconception": "you subtracted the actual yield from the theoretical yield and reported the difference as a percent",
"feedback": "This answer is wrong — you subtracted the actual yield from the theoretical yield and reported the difference as a percent."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nactual yield = 13.3 g\ntheoretical yield = 76.0 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nSubstitute in the values, and calculate:\npercent yield = (13.3 ÷ 76) × 100\npercent yield = 17.5%\nThe actual yield is far below the theoretical yield, so the percentage must come out well below 100% — an answer like 571% fails that check."
}
STO-027 · medium · application · seq 11 · authored (fresh pair 275 g at 28% -> exact zero-free 77 g; '28' ≠ corpus '28.0' and '77' ≠ corpus '0.63' string-forms; seed p12's structure and diagnostic classes kept; tail = whitelisted convention phrase) · numeric
A scaled-up preparation run has a theoretical yield of 275 g, and the process reliably achieves a 28% yield. What mass of product should the run expect to collect? Give your answer in grams to 2 significant figures.
Answer: 77 g KEY
also accepted: 77 g, 77
wrongValueDiagnostics
[
{
"value": 982.1,
"misconception": "you divided the theoretical yield by the percent fraction instead of multiplying — a below-100% yield must give an actual yield below the theoretical 275 g",
"feedback": "This answer is wrong — you divided the theoretical yield by the percent fraction instead of multiplying — a below-100% yield must give an actual yield below the theoretical 275 g."
},
{
"value": 7700,
"misconception": "you forgot to divide the percent yield by 100 — thousands of grams from a 275 g prediction is the signal",
"feedback": "This answer is wrong — you forgot to divide the percent yield by 100 — thousands of grams from a 275 g prediction is the signal."
},
{
"value": 247,
"misconception": "you subtracted the percent number from the theoretical yield as if it were a mass in grams — 28 is a percentage, not a mass",
"feedback": "This answer is wrong — you subtracted the percent number from the theoretical yield as if it were a mass in grams — 28 is a percentage, not a mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\npercent yield = 28%\ntheoretical yield = 275 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nMake the actual yield the subject:\nactual yield = (percent yield ÷ 100) × theoretical yield\nSubstitute in the values, and calculate:\nactual yield = (28 ÷ 100) × 275\nactual yield = 77 g\nSanity check: 77 g is below the theoretical 275 g — a percent yield below 100% must give an actual yield below the theoretical yield."
}
STO-028 · medium · application · seq 12 · authored (fresh pair 94/106, both strings prescan-clean; over-cap 106 >= 103; notebook two-trial surface; seed p15's structure and distractor classes kept; nine-tenths word-dressing keeps the invented cap below 94 without a numeral) · mcq4
A notebook records two trials of one synthesis: trial 1 lists a 94% yield, and trial 2 lists a 106% yield. Which trial's value needs rechecking?
Trial 1's — 94% is suspiciously close to perfect for a real synthesis.
Both trials' — a real run cannot keep even nine-tenths of its predicted product.
Neither trial's — values on either side of 100% are equally normal outcomes.
STO-026 · medium · application · seq 13 · authored (fresh exact-division dataset pair 440/209 vs 275/143 -> 47.5% vs 52%, winner second-listed with fewer grams; close-results variant — the 8-point gap forces both computations; '47.5' ≠ corpus '44.0' and '143' ≠ corpus '142.9' string-form exactness noted; seed p11's structure and distractor classes kept; results-only options) · mcq4
A lab team evaluates two reactions. Reaction 1 has a theoretical yield of 440 g and collects 209 g of product. Reaction 2 has a theoretical yield of 275 g and collects 143 g. Which reaction has the higher percent yield?
Reaction 2 — 52% compared with 47.5%. KEY
Reaction 1 — more grams collected always means a higher percent yield.
Reaction 1 — 211% compared with 192%.
The two reactions tie — a miss is a miss, no matter how large the gap below the prediction.
STO-029 · medium · application · seq 14 · authored (fresh combustion family 2Ni + O₂ -> 2NiO with the 2:2 coefficients exercising the ratio step explicitly; excess convention per A-U8G3-2; exact route 0.800 mol -> T 59.76 g -> 7/8 = 87.5%; all stem strings ledger-free, 58.7 ≠ corpus 55.8 string-exact) · numeric
46.96 g of nickel powder burns in excess O₂ by 2Ni + O₂ → 2NiO, and 52.29 g of NiO is collected. What is the percent yield? (M of Ni = 58.7 g/mol; M of NiO = 74.7 g/mol.) Give your answer as a percentage to 3 significant figures.
Answer: 87.5 % KEY
also accepted: 87.5%
wrongValueDiagnostics
[
{
"value": 111,
"misconception": "you divided the product mass by the reactant mass — the comparison is with the theoretical yield, 59.76 g; comparing with the metal's mass even pushes the result past 100, which a genuine yield can never do",
"feedback": "This answer is wrong — you divided the product mass by the reactant mass — the comparison is with the theoretical yield, 59.76 g; comparing with the metal's mass even pushes the result past 100, which a genuine yield can never do."
},
{
"value": 114,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
}
]
feedback
{
"keyExplanation": "Step 1 — convert the reactant mass to moles:\nn(Ni) = 46.96 / 58.7 = 0.800 mol\nStep 2 — convert to moles of product with the mole ratio:\nmoles of NiO = 0.800 × (2 ÷ 2) = 0.800 mol\nStep 3 — convert the product moles to mass (the theoretical yield):\ntheoretical yield = 0.800 × 74.7 = 59.76 g\nStep 4 — calculate the percent yield:\npercent yield = (52.29 ÷ 59.76) × 100\npercent yield = 87.5%"
}
STO-030 · hard · application · seq 15 · authored (fresh taught-class precipitation: manganese(II) sulfate + lithium carbonate → MnCO₃, 1:1; the excess Li₂CO₃ partner works as an excess solid — it keeps dissolving as the far less soluble MnCO₃ precipitates; 'sulfate solution is stirred with' dodges fluency-3's 'sulfate solution reacts with' 4-gram; no material repeats within form D — row09-D is nickel/K₃PO₄ (manganese sits in row09's form C and row14's form B, different tests) and row14-D is nickel) | 2026-09-08 crash-resume: values re-freshened 620 mL / 0.15 M → 550 mL / 1.6 M ('0.15' collided cross-form with B#9 and C#15; '1.5'/'620'-successor candidates sat in spec text or the inventory; '550' was freed by the C#13 re-value this same wave), answer re-derived 0.88 mol (MnSO₄ stays comfortably inside its ~3 M solubility); diagnostics and scaffold re-derived. · numeric
550 mL of 1.6 M manganese(II) sulfate solution is stirred with excess lithium carbonate. The reaction is MnSO₄ + Li₂CO₃ → Li₂SO₄ + MnCO₃. What amount of MnCO₃, in moles, forms? Give your answer in moles to 2 significant figures.
Answer: 0.88 mol KEY
also accepted: 0.88 mol
wrongValueDiagnostics
[
{
"value": 880.0,
"misconception": "you used the volume in milliliters — n = M × V needs liters, and 550 mL is 0.550 L",
"feedback": "This answer is wrong — you used the volume in milliliters — n = M × V needs liters, and 550 mL is 0.550 L."
},
{
"value": 1.6,
"misconception": "you used the molarity as the mole amount without multiplying by the volume",
"feedback": "This answer is wrong — you used the molarity as the mole amount without multiplying by the volume."
},
{
"value": 0.34,
"misconception": "you divided the volume by the molarity instead of multiplying them",
"feedback": "This answer is wrong — you divided the volume by the molarity instead of multiplying them."
}
]
feedback
{
"keyExplanation": "Convert the volume to liters first: 550 mL = 0.550 L.\nStep 1 — moles of dissolved reactant:\nn(MnSO₄) = M × V\nn(MnSO₄) = 1.6 × 0.550 = 0.88 mol\nStep 2 — moles of product with the mole ratio:\nmoles of MnCO₃ = 0.88 × (1 ÷ 1)\nmoles of MnCO₃ = 0.88 mol\nA 880 mol answer fails the sanity check — just over half a liter of solution holds around a mole of solute at this concentration, not hundreds."
}
STO-029 · hard · application · seq 16 · authored (fresh selenium combustion — the seed's same-group chemistry with a fresh element and data; near-coincidence trap: collected 7.881 g vs 7.90 g burned, naive comparison 99.8% vs true 71.0%; Se 79.0 / SeO₂ 111.0 real at 1 decimal and internally consistent; exact route 0.100 mol -> 11.10 g -> 71.0%; masses ≤12 unscanned, molar-mass strings ledger-free — '111.0' ≠ corpus '112') · numeric
7.90 g of selenium burns in excess O₂ by Se + O₂ → SeO₂, and 7.881 g of SeO₂ is collected. What is the percent yield? (M of Se = 79.0 g/mol; M of SeO₂ = 111.0 g/mol.) Give your answer as a percentage to 3 significant figures.
Answer: 71.0 % KEY
also accepted: 71.0%
wrongValueDiagnostics
[
{
"value": 140.8,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 99.8,
"misconception": "you compared the product mass with the reactant mass — the theoretical yield is 11.10 g of SeO₂, not 7.90 g",
"feedback": "This answer is wrong — you compared the product mass with the reactant mass — the theoretical yield is 11.10 g of SeO₂, not 7.90 g."
},
{
"value": 0.71,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
}
]
feedback
{
"keyExplanation": "Step 1 — convert the reactant mass to moles:\nn(Se) = 7.90 / 79.0 = 0.100 mol\nStep 2 — convert to moles of product with the mole ratio:\nmoles of SeO₂ = 0.100 × (1 ÷ 1) = 0.100 mol\nStep 3 — convert the product moles to mass (the theoretical yield):\ntheoretical yield = 0.100 × 111.0 = 11.10 g\nStep 4 — calculate the percent yield:\npercent yield = (7.881 ÷ 11.10) × 100\npercent yield = 71.0%\nThe near-match between 7.881 g of product and 7.90 g of reactant is a coincidence of masses — the yield comparison is always against the theoretical yield."
}
STO-024 · hard · transfer · seq 17 · authored (gas-mass-vanished-on-cooling claim rotation per blueprint — spare p5's distractor class elevated to the evaluated claim; chilled-collection-vessel surface; cooling-changes-volume-not-mass key with the escaped-gas transfer destination scenario-fitted; gap mass 2 g <= 12; instance-free) · mcq4
A reaction's gaseous product is piped into a chilled collection vessel and weighed once collection ends. The actual yield weighs in 2 g below the theoretical yield. A classmate claims the gas simply lost mass as it cooled in the vessel. What is the correct account?
Cooling changes a gas's volume, not its mass — the missing 2 g is matter that ended up elsewhere, such as unreacted material, side products, or gas that never reached the vessel. KEY
The classmate is right — a gas weighs less once it has cooled and contracted.
Some atoms of the gas were destroyed by the sudden temperature drop.
The missing mass was carried away as the heat the gas gave up while cooling — when energy leaves a sample, matter must leave with it, gram for gram, in equal measure.
STO-027 · hard · transfer · seq 18 · authored (fresh triple: 600 g at 39% -> expected 234 g vs 245 g contracted, exact, all strings prescan-clean; complement 366 g coherent above the contract since 39% < 50; delivery-contract surface; seed p14's structure, No verdict, key position, and all three distractor classes kept) · mcq4
A team is contracted to deliver 245 g of a compound. The planned run has a theoretical yield of 600 g, and the plant typically gives a 39% yield. Is the run expected to meet the contract?
No — the expected actual yield is 234 g, below the 245 g contracted. KEY
Yes — a 600 g theoretical yield makes the 245 g contracted a sure thing.
Yes — the expected actual yield is 366 g, comfortably above the 245 g contracted.
It cannot be judged — finding the expected mass requires the molar masses of the reactants and the compound.
STO-028 · hard · transfer · seq 19 · authored (extra-of-every-reactant dressing per blueprint; key and explanation state that the recalculated theoretical yield travels UP with the loading — the seed's limiting-reactant account applied to the trickiest dressing, recorded in authoring_notes so no fix loop inverts it; numeral surface = 100 only; instance-free) · mcq4
A group reasons: 'If we load a generous extra amount of every reactant, nothing can run short, so the percent yield must come out above 100%.' What is wrong with the reasoning?
The theoretical yield is recalculated from the loaded amounts and still sets the run's maximum product — the collected amount cannot rise above it. KEY
Nothing is wrong — with every reactant in generous supply, real runs really do pass a 100% yield.
The reasoning fails because percent yield is set by how fast the reaction runs, and the extra loading merely speeds the reaction up.
The reasoning fails because adding more of every reactant drags the theoretical yield down instead.
STO-029 · hard · transfer · seq 20 · authored (fresh metal-oxidation pair: copper heated in excess O₂ to black copper(II) oxide, 1:1; 'copper' also appears in this same form at D#6 (the crash-resume rewrite) — same-form non-adjacent seqs 6/20, which the validator routes to a telegraphing WARN for blind review, recorded; cross-form copper is unique to form D; cross-lesson WARN vs STO-030 recorded; all values spec- and inventory-clean; raw-grams mislead holds (13.674 > 9.381 while 43% < 59%)) · mcq4
Two runs of 2Cu + O₂ → 2CuO heat different amounts of copper in excess O₂ until the black oxide forms (M of Cu = 63.5 g/mol; M of CuO = 79.5 g/mol). Run 1: 25.4 g of Cu, 13.674 g of CuO collected. Run 2: 12.7 g of Cu, 9.381 g of CuO collected. Which run has the higher percent yield?
Run 2 — 59% compared with 43%. KEY
Run 1 — collecting more grams of CuO (13.674 g against 9.381 g) shows it had the higher percent yield.
Run 1 — 233% compared with 169%.
The runs tie — identical chemistry means identical percent yields, however large the run.
STO-030 · medium · application · seq 21 · authored (FRQ-1 form D: fresh taught-class precipitation MnCl₂ + 2KOH → 2KCl + Mn(OH)₂, 1 : 1 asked ratio with the 2KOH coefficient as the classic ratio lure; Mn(OH)₂ is not amphoteric so the excess KOH cannot redissolve it; manganese sits in a different form than seq 9's manganese; exact product 0.0988 mol → 0.099 mol at 2 s.f.) · frq
0.52 L of 0.19 M MnCl₂ solution is treated with excess potassium hydroxide by MnCl₂ + 2KOH → 2KCl + Mn(OH)₂.
0.52 L of 0.19 M MnCl₂ solution is treated with excess potassium hydroxide by MnCl₂ + 2KOH → 2KCl + Mn(OH)₂.
(a) Calculate how many moles of MnCl₂ are dissolved in the 0.52 L of solution. Show your working.
(b) Use the mole ratio from the balanced equation to calculate how many moles of Mn(OH)₂ form. Show your working. Give your answer in moles to 2 significant figures.
Part a. Calculate how many moles of MnCl₂ are dissolved in the 0.52 L of solution. Show your working.
Model: Values: V = 0.52 L of solution, M = 0.19 mol/L — M is the molarity of the manganese(II) chloride solution, not a molar mass.
n = M × V
n = 0.19 × 0.52
n = 0.0988 mol of manganese(II) chloride dissolved.
Part b. Use the mole ratio from the balanced equation to calculate how many moles of Mn(OH)₂ form. Show your working. Give your answer in moles to 2 significant figures.
Model: From the balanced equation, 1 mol of manganese(II) hydroxide forms for every 1 mol of manganese(II) chloride.
moles of manganese(II) hydroxide = 0.0988 × (1 ÷ 1)
moles of manganese(II) hydroxide = 0.0988 mol
To 2 significant figures: 0.099 mol of manganese(II) hydroxide. The product stays in moles.
Full model answer. (a) Values: V = 0.52 L of solution, M = 0.19 mol/L — M is the molarity of the manganese(II) chloride solution, not a molar mass.
n = M × V
n = 0.19 × 0.52
n = 0.0988 mol of manganese(II) chloride dissolved.
(b) From the balanced equation, 1 mol of manganese(II) hydroxide forms for every 1 mol of manganese(II) chloride.
moles of manganese(II) hydroxide = 0.0988 × (1 ÷ 1)
moles of manganese(II) hydroxide = 0.0988 mol
To 2 significant figures: 0.099 mol of manganese(II) hydroxide. The product stays in moles.
STO-029 · hard · application · seq 22 · authored (FRQ-2 form D: lithium burn, 4:2 ratio; low-yield register exercises the below-100 sanity frame; UNSEEDED) · frq
A strip of lithium with a mass of 20.7 g burns in excess O₂ by 4Li + O₂ → 2Li₂O. After cooling, 8.97 g of Li₂O is collected.
A strip of lithium with a mass of 20.7 g burns in excess O₂ by 4Li + O₂ → 2Li₂O. After cooling, 8.97 g of Li₂O is collected.
(a) Calculate the theoretical yield of Li₂O. Show every step of your working. Report the theoretical yield to 3 significant figures.
(b) Calculate the percent yield of Li₂O. Give your answer as a percentage to 3 significant figures.
Molar masses (exact): Li 6.9 g/mol, Li₂O 29.9 g/mol.
Write your answers in plain numbers and words; no special symbols are needed.
Part a. Calculate the theoretical yield of Li₂O. Show every step of your working. Report the theoretical yield to 3 significant figures.
Model: Values: mass of lithium = 20.7 g; molar mass of lithium = 6.9 g/mol; molar mass of lithium oxide = 29.9 g/mol.
Equation: n = m ÷ M
Substitute: n = 20.7 ÷ 6.9
Calculate: n = 3.00 mol of lithium
Mole ratio from the balanced equation: 2 mol of lithium oxide forms from 4 mol of lithium, so n = 3.00 × (2 ÷ 4) = 1.50 mol of lithium oxide
Equation: m = n × M
Substitute: m = 1.50 × 29.9
Calculate: theoretical yield = 44.85 g, reported as 44.9 g to 3 significant figures
Part b. Calculate the percent yield of Li₂O. Give your answer as a percentage to 3 significant figures.
Molar masses (exact): Li 6.9 g/mol, Li₂O 29.9 g/mol.
Write your answers in plain numbers and words; no special symbols are needed.
Model: Values: actual yield = 8.97 g; theoretical yield = 44.85 g.
Equation: percent yield = (actual yield ÷ theoretical yield) × 100
Substitute: percent yield = (8.97 ÷ 44.85) × 100
Calculate: percent yield = 20.0% (the rounded 44.9 g gives the same 20.0%)
Full model answer. (a) Values: mass of lithium = 20.7 g; molar mass of lithium = 6.9 g/mol; molar mass of lithium oxide = 29.9 g/mol.
Equation: n = m ÷ M
Substitute: n = 20.7 ÷ 6.9
Calculate: n = 3.00 mol of lithium
Mole ratio from the balanced equation: 2 mol of lithium oxide forms from 4 mol of lithium, so n = 3.00 × (2 ÷ 4) = 1.50 mol of lithium oxide
Equation: m = n × M
Substitute: m = 1.50 × 29.9
Calculate: theoretical yield = 44.85 g, reported as 44.9 g to 3 significant figures
(b) Values: actual yield = 8.97 g; theoretical yield = 44.85 g.
Equation: percent yield = (actual yield ÷ theoretical yield) × 100
Substitute: percent yield = (8.97 ÷ 44.85) × 100
Calculate: percent yield = 20.0% (the rounded 44.9 g gives the same 20.0%)
A reaction was run last week, and its product was weighed and stored. Only afterwards does a chemist work through the stoichiometry calculation, getting 5.8 g as the amount the reaction could have produced. What is the 5.8 g?
The theoretical yield. KEY
The actual yield.
Both yields at once.
Neither — a calculation done after the run comes too late to be a yield.
STO-025 · easy · recall · seq 2 · authored (recall probe E: what the x-100 step does; factor spelled 'one hundred' in words — no digit '100' in stem/options, keeping row 2 off the taught-constant dependency list; '(actual yield ÷ theoretical yield)' is rule-line-2 text, A3 whitelist source) · mcq4
The last step of the percent yield equation multiplies the fraction (actual yield ÷ theoretical yield) by one hundred. What does this multiplication do?
It rescales the unitless fraction into a percentage. KEY
It converts the yields from grams into a larger mass unit.
It multiplies the actual yield itself, making the measured mass larger.
It cancels the gram units shared by the two yields.
A percent yield of exactly 100% means the collected product exactly matched one particular quantity. Which quantity?
The theoretical yield — the calculated most product the reactants can possibly make. KEY
The total mass of reactants weighed out at the start of the run.
The largest yield any group has ever recorded for that reaction.
The combined mass of everything in the flask when the reaction stops, product and leftovers together.
STO-023 · easy · application · seq 4 · authored (surface: synthesis record sheet — fourth written-record surface added to the blueprint's three; probe: actual yield; values 82/73 ledger-verified) · mcq4
A synthesis record sheet shows two entries: 'target mass from calculation: 82 g' and 'product mass on the balance: 73 g'. Which entry is the actual yield?
The 73 g, because it was read off the balance after the reaction. KEY
The 82 g, because the target set by the calculation is the more official number.
Both entries, because the sheet records them for one product.
Neither entry — the actual yield is the 9 g lost between calculation and weighing.
A lab poster shows the equation percent yield = actual yield × theoretical yield × 100. A student says the poster is wrong and writes percent yield = (actual yield ÷ theoretical yield) × 100. Who is right?
The student — percent yield divides the actual yield by the theoretical yield. KEY
The poster — multiplying the two yields combines the measurement and the prediction into one number.
Both — multiplying and dividing the yields give the same comparison once the units cancel.
Neither — both versions should drop the × 100 step.
STO-024 · medium · application · seq 6 · authored (transfer-loss rotation: splashed-transfer surface per blueprint; distinct handling surface from form B — liquid splash vs solid residue; instance-light) | blind-wave1 2026-09-08: evidence-scoped ask (F2 class sweep) · mcq4
While pouring a liquid product from the reaction flask into a beaker for weighing, a student splashes some of it onto the bench. The measured actual yield ends up below the theoretical yield. Which cause does this observation directly show?
Product was lost on transfer. KEY
A side reaction consumed reactant.
The reaction stopped before completion.
Atoms in the splashed liquid were destroyed on contact with the bench.
STO-025 · medium · application · seq 7 · authored (fresh exact-division yield pair 92/46, ratio 1:2, both strings prescan-clean; experiment surface; seed p6's structure and distractor classes kept) · mcq4
For one experiment, the theoretical yield is 92 g and the actual yield is 46 g. Which substitution into the percent yield equation is correct?
(46 ÷ 92), multiplied by 100 KEY
(92 ÷ 46), multiplied by 100
(46 ÷ 92), divided by 100
(92 − 46), read directly as a percentage
STO-026 · medium · application · seq 8 · authored (high-yield 94.0%; demonstration surface; values 30.0/28.2 ledger-free string-form — corpus carries '30'/'0.30', not '30.0'; exact division) · numeric
A demonstration reaction has a theoretical yield of 30.0 g. The teacher collects 28.2 g of product. What is the percent yield? Give your answer as a percentage to 3 significant figures.
Answer: 94.0 % KEY
also accepted: 94.0%
wrongValueDiagnostics
[
{
"value": 106,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 0.94,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
},
{
"value": 1.8,
"misconception": "you subtracted the yields — percent yield compares them by division, not by difference",
"feedback": "This answer is wrong — you subtracted the yields — percent yield compares them by division, not by difference."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nactual yield = 28.2 g\ntheoretical yield = 30.0 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nSubstitute in the values, and calculate:\npercent yield = (28.2 ÷ 30.0) × 100\npercent yield = 94.0%"
}
STO-030 · medium · application · seq 9 · authored (parallel numeric slot; fresh taught-class precipitation CoCl₂ + K₂CO₃ → CoCO₃ + 2KCl, 1 : 1 asked ratio with the 2KCl coefficient as the doubling lure; values 0.45 L / 0.18 M ledger-free, exact product) · numeric
0.45 L of 0.18 M cobalt(II) chloride solution reacts with excess potassium carbonate by CoCl₂ + K₂CO₃ → CoCO₃ + 2KCl. How many moles of CoCO₃ form? Give your answer in moles to 2 significant figures.
Answer: 0.081 mol KEY
also accepted: 0.081 mol
wrongValueDiagnostics
[
{
"value": 0.18,
"misconception": "you used the molarity as the mole amount without multiplying by the volume",
"feedback": "This answer is wrong — you used the molarity as the mole amount without multiplying by the volume."
},
{
"value": 0.16,
"misconception": "you doubled the moles with KCl's coefficient — CoCO₃'s ratio to CoCl₂ is 1 : 1",
"feedback": "This answer is wrong — you doubled the moles with KCl's coefficient — CoCO₃'s ratio to CoCl₂ is 1 : 1."
}
]
feedback
{
"keyExplanation": "Step 1 — moles of dissolved reactant:\nn(CoCl₂) = M × V\nn(CoCl₂) = 0.18 × 0.45 = 0.081 mol\nStep 2 — moles of product with the mole ratio:\nmoles of CoCO₃ = 0.081 × (1 ÷ 1)\nmoles of CoCO₃ = 0.081 mol"
}
STO-026 · medium · application · seq 10 · authored (fresh low-yield pair 47 g / 200 g -> 23.5%; strings '47'/'200'/'23.5' ledger-free ('200' != corpus '0.200'); exact division 47/200 = 0.235; answer not a bare power of ten; 200 g = demonstration-scale framing, not a round-number trap) · numeric
A demonstration-scale reaction has a theoretical yield of 200 g. The experiment collects 47 g of product. What is the percent yield? Give your answer as a percentage to 3 significant figures.
Answer: 23.5 % KEY
also accepted: 23.5%
wrongValueDiagnostics
[
{
"value": 426,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 0.235,
"misconception": "you forgot to multiply the fraction by 100",
"feedback": "This answer is wrong — you forgot to multiply the fraction by 100."
},
{
"value": 153,
"misconception": "you subtracted the actual yield from the theoretical yield and reported the difference as a percent",
"feedback": "This answer is wrong — you subtracted the actual yield from the theoretical yield and reported the difference as a percent."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nactual yield = 47 g\ntheoretical yield = 200 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nSubstitute in the values, and calculate:\npercent yield = (47 ÷ 200) × 100\npercent yield = 23.5%\nThe actual yield is far below the theoretical yield, so the percentage must come out well below 100% — an answer like 426% fails that check."
}
STO-027 · medium · application · seq 11 · authored (fresh pair 675 g at 12% -> exact zero-free 81 g; the 12 percent sits in the scan-ignored 0..12 band; divided-diagnostic 5625 is exact; seed p12's structure and diagnostic classes kept; tail = whitelisted convention phrase) · numeric
An industrial run has a theoretical yield of 675 g, and the process reliably gives a 12% yield. What mass of product should the run expect to collect? Give your answer in grams to 2 significant figures.
Answer: 81 g KEY
also accepted: 81 g
wrongValueDiagnostics
[
{
"value": 5600,
"misconception": "you divided the theoretical yield by the percent fraction instead of multiplying — a below-100% yield must give an actual yield below the theoretical 675 g",
"feedback": "This answer is wrong — you divided the theoretical yield by the percent fraction instead of multiplying — a below-100% yield must give an actual yield below the theoretical 675 g."
},
{
"value": 8100,
"misconception": "you forgot to divide the percent yield by 100 — thousands of grams from a 675 g prediction is the signal",
"feedback": "This answer is wrong — you forgot to divide the percent yield by 100 — thousands of grams from a 675 g prediction is the signal."
},
{
"value": 663,
"misconception": "you subtracted the percent number from the theoretical yield as if it were a mass in grams — 12 is a percentage, not a mass",
"feedback": "This answer is wrong — you subtracted the percent number from the theoretical yield as if it were a mass in grams — 12 is a percentage, not a mass."
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\npercent yield = 12%\ntheoretical yield = 675 g\nWrite down the equation:\npercent yield = (actual yield ÷ theoretical yield) × 100\nMake the actual yield the subject:\nactual yield = (percent yield ÷ 100) × theoretical yield\nSubstitute in the values, and calculate:\nactual yield = (12 ÷ 100) × 675\nactual yield = 81 g\nSanity check: 81 g is below the theoretical 675 g — a percent yield below 100% must give an actual yield below the theoretical yield."
}
STO-028 · medium · application · seq 12 · authored (fresh pair 89/121, both strings prescan-clean; over-cap 121 >= 103; handed-in-values surface with the pair described positionally — no group labels needed; seed p15's structure and distractor classes kept; word-dressed cap distractors) · mcq4
Two student groups hand in yield values for the same procedure: 89% from one group and 121% from the other. Which value needs rechecking?
The 121% value — a genuine percent yield cannot rise above 100%. KEY
The 89% value — losing so little product is too good to be honest.
Both values — genuine student yields never climb into the high eighties, let alone higher.
Neither value — percent yields swing well above and below 100% from run to run.
STO-026 · medium · application · seq 13 · authored (fresh exact-division dataset pair 850/357 vs 400/308 -> 42% vs 77%, winner first-listed with fewer grams; high-yield-register variant on both sides; synthesis-route surface; all strings prescan-clean; seed p11's structure and distractor classes kept; results-only options) · mcq4
Two synthesis routes to the same product are compared. Route 1 has a theoretical yield of 850 g and collects 357 g. Route 2 has a theoretical yield of 400 g and collects 308 g. Which route has the higher percent yield?
Route 2 — 77% compared with 42%. KEY
Route 1 — the route that collects more grams always has the higher percent yield.
Route 1 — 238% compared with 130%.
The two routes tie — collecting less than predicted counts the same however large the gap is.
STO-029 · medium · application · seq 14 · authored (fresh combustion family 4Cr + 3O₂ -> 2Cr₂O₃ — the row's non-1:1 variant, so the skipped-ratio diagnostic is live here and only here; chromium(III) oxide, not hexavalent, classroom-safe; excess convention per A-U8G3-2; exact route 0.400 mol -> ratio 2:4 -> 0.200 mol -> T 30.40 g -> 17/20 = 85.0%; all stem strings ledger-free) · numeric
20.8 g of chromium powder burns in excess O₂ by 4Cr + 3O₂ → 2Cr₂O₃, and 25.84 g of Cr₂O₃ is collected. What is the percent yield? (M of Cr = 52.0 g/mol; M of Cr₂O₃ = 152.0 g/mol.) Give your answer as a percentage to 3 significant figures.
Answer: 85.0 % KEY
also accepted: 85.0%
wrongValueDiagnostics
[
{
"value": 42.5,
"misconception": "you skipped the mole-ratio step — 0.400 mol is the chromium's moles, and every 4 mol of Cr makes only 2 mol of Cr₂O₃",
"feedback": "This answer is wrong — you skipped the mole-ratio step — 0.400 mol is the chromium's moles, and every 4 mol of Cr makes only 2 mol of Cr₂O₃."
},
{
"value": 118,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 124,
"misconception": "you divided the product mass by the reactant mass — the comparison is with the theoretical yield, 30.40 g; comparing with the metal's mass even pushes the result past 100, which a genuine yield can never do",
"feedback": "This answer is wrong — you divided the product mass by the reactant mass — the comparison is with the theoretical yield, 30.40 g; comparing with the metal's mass even pushes the result past 100, which a genuine yield can never do."
}
]
feedback
{
"keyExplanation": "Step 1 — convert the reactant mass to moles:\nn(Cr) = 20.8 / 52.0 = 0.400 mol\nStep 2 — convert to moles of product with the mole ratio:\nmoles of Cr₂O₃ = 0.400 × (2 ÷ 4) = 0.200 mol\nStep 3 — convert the product moles to mass (the theoretical yield):\ntheoretical yield = 0.200 × 152.0 = 30.40 g\nStep 4 — calculate the percent yield:\npercent yield = (25.84 ÷ 30.40) × 100\npercent yield = 85.0%"
}
STO-030 · hard · application · seq 15 · authored (2026-09-08 crash-resume rebuild: the shipped form E used tin + 750 mL — 'tin' collided cross-form with B#9 and D#6, and '750' with A#15's volume. Rebuilt as the taught METAL–ACID class with manganese: Mn dissolves readily in dilute H₂SO₄ to MnSO₄ + H₂ (genuine, classroom-real), 1:1 to the acid, acid is the dissolved limiting reactant with the metal in stated excess; 'manganese' is non-lexicon; within form E no other manganese item exists (E#9 cobalt, E#14 chromium); volume 500 mL / M kept at 0.92; mL trap, three diagnostic classes, scaffold and sanity line kept) · numeric
A beaker holds 500 mL of 0.92 M dilute sulfuric acid. Excess manganese is added, and the metal dissolves as it reacts by Mn + H₂SO₄ → MnSO₄ + H₂. How many moles of MnSO₄ form? Give your answer in moles to 2 significant figures.
Answer: 0.46 mol KEY
also accepted: 0.46 mol
wrongValueDiagnostics
[
{
"value": 460.0,
"misconception": "you used the volume in milliliters — n = M × V needs liters, and 500 mL is 0.500 L",
"feedback": "This answer is wrong — you used the volume in milliliters — n = M × V needs liters, and 500 mL is 0.500 L."
},
{
"value": 0.92,
"misconception": "you used the molarity as the mole amount without multiplying by the volume",
"feedback": "This answer is wrong — you used the molarity as the mole amount without multiplying by the volume."
},
{
"value": 0.54,
"misconception": "you divided the volume by the molarity instead of multiplying them",
"feedback": "This answer is wrong — you divided the volume by the molarity instead of multiplying them."
}
]
feedback
{
"keyExplanation": "Convert the volume to liters first: 500 mL = 0.500 L.\nStep 1 — moles of dissolved reactant:\nn(H₂SO₄) = M × V\nn(H₂SO₄) = 0.92 × 0.500 = 0.46 mol\nStep 2 — moles of product with the mole ratio:\nmoles of MnSO₄ = 0.46 × (1 ÷ 1)\nmoles of MnSO₄ = 0.46 mol\nA 460 mol answer fails the sanity check — half a liter of a dilute acid holds well under a mole of solute."
}
STO-029 · hard · application · seq 16 · authored (fresh boron combustion; tempting pattern: collected 6.09 g EXCEEDS the 5.40 g of boron burned — naive comparison 112.8% looks impossible and forces the theoretical-yield route; true 35.0%; 4:2 ratio rotates the mole-ratio step live (seq-9 rotation precedent, step count unchanged) and enables the specNotes skipped-ratio diagnostic (17.5); B 10.8 / B₂O₃ 69.6 real at 1 decimal and internally consistent (2×10.8 + 3×16.0); exact route 0.500 mol -> 0.250 mol -> 17.4 g -> 35.0%; masses ≤12 unscanned, '69.6' ledger-free) · numeric
5.40 g of boron burns in excess O₂ by 4B + 3O₂ → 2B₂O₃, and 6.09 g of B₂O₃ is collected. What is the percent yield? (M of B = 10.8 g/mol; M of B₂O₃ = 69.6 g/mol.) Give your answer as a percentage to 3 significant figures.
Answer: 35.0 % KEY
also accepted: 35.0%
wrongValueDiagnostics
[
{
"value": 286,
"misconception": "you divided the theoretical yield by the actual yield — the measured amount goes on top",
"feedback": "This answer is wrong — you divided the theoretical yield by the actual yield — the measured amount goes on top."
},
{
"value": 113,
"misconception": "you compared the product mass with the reactant mass — a collected mass above the boron mass is possible because the oxide gains oxygen; the theoretical yield is 17.4 g of B₂O₃",
"feedback": "This answer is wrong — you compared the product mass with the reactant mass — a collected mass above the boron mass is possible because the oxide gains oxygen; the theoretical yield is 17.4 g of B₂O₃."
},
{
"value": 17.5,
"misconception": "you skipped the mole ratio — every 4 mol of B makes only 2 mol of B₂O₃",
"feedback": "This answer is wrong — you skipped the mole ratio — every 4 mol of B makes only 2 mol of B₂O₃."
}
]
feedback
{
"keyExplanation": "Step 1 — convert the reactant mass to moles:\nn(B) = 5.40 / 10.8 = 0.500 mol\nStep 2 — convert to moles of product with the mole ratio:\nmoles of B₂O₃ = 0.500 × (2 ÷ 4) = 0.250 mol\nStep 3 — convert the product moles to mass (the theoretical yield):\ntheoretical yield = 0.250 × 69.6 = 17.4 g\nStep 4 — calculate the percent yield:\npercent yield = (6.09 ÷ 17.4) × 100\npercent yield = 35.0%\nCollecting more grams of B₂O₃ (6.09 g) than the boron burned (5.40 g) is not a contradiction — the oxide carries added oxygen mass — and the yield comparison is always against the theoretical yield."
}
STO-024 · hard · transfer · seq 17 · authored (atoms-destroyed claim in a fresh dressing per blueprint; open-crucible strong-heating surface, spatter/carried-out destination scenario-fitted; four distinct accounts — conservation key + endorse + matter-to-heat + double-counting; gap mass 5 g <= 12; instance-free) · mcq4
After strong heating in an open crucible, a run's actual yield comes to 5 g less than the theoretical yield. A bench partner insists that the fierce heating destroyed 5 g of atoms outright. What is the correct account?
No atoms were destroyed — the missing mass is matter that ended up elsewhere, such as unreacted material, side products, or material carried out of the open crucible. KEY
The partner is right — fierce enough heating breaks atoms down to nothing.
The missing mass became the glow and heat the crucible gave off.
The theoretical yield counted the crucible's contents twice, so the prediction started out too high and the run never had a chance to reach it.
STO-027 · hard · transfer · seq 18 · authored (VERDICT ROTATED to deliver with a tight margin: 650 g at 58% -> expected 377 g vs 370 g ordered, exact, all strings prescan-clean; the 7 g margin forces the computation; percent > 50 puts the complement 273 g below the order so the flipped-percentage option carries the wrong No verdict; compared-to-theoretical class dressed as the full-collection claim; firm-order surface; two deliver-verdict options recorded for the blind round) · mcq4
A specialty chemicals firm accepts an order for 370 g of a compound. The planned run has a theoretical yield of 650 g, and the reaction typically gives a 58% yield. Is the run expected to fill the order?
Yes — the expected actual yield is 377 g, just above the 370 g ordered. KEY
Yes — the run will collect the full 650 g planned, covering the order nearly twice over.
No — the expected actual yield is only 273 g, well short of the 370 g ordered.
It cannot be judged — an expected mass cannot be found without the molar masses of the compound and the reactants.
STO-028 · hard · transfer · seq 19 · authored (industrial-scale-flooding dressing per blueprint; 'production line'/'batch' vocabulary — the word 'reactor' deliberately avoided per the course-wide banned-framings rule; seed p16's construct, key account, and all three wrong-account classes kept; numeral surface = 100 only; instance-free) · mcq4
A brochure claims: 'Our production line beats 100% yields by flooding every batch with a huge excess of the cheaper reactant.' What is wrong with the claim?
The theoretical yield already counts all of the limiting reactant — no scale of excess can lift the collected product above it. KEY
Nothing is wrong — at production scale, flooding a batch with a large excess genuinely lifts the collected product beyond a 100% yield.
The claim fails only because a huge excess slows each batch down too much to be profitable.
The claim fails because flooding a batch with excess shrinks its theoretical yield instead.
STO-029 · hard · transfer · seq 20 · authored (fresh decomposition with the taught 2:1 ratio: baking-soda decomposition 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂, classroom classic, product = residue weighed on the balance; 'sodium' = cross-lesson WARN class (STO-030 fluency mat) recorded — it is NOT an STO-029 own material (own set: carbon/iron/magnesium/methane/water, all avoided) and no other form carries sodium; the 2:1 ratio exercises the full mole-route both runs identically (parallel structure within the item preserved); all values spec- and inventory-clean; raw-grams mislead holds (7.685 > 7.526 while 29% < 71%)) · mcq4
Two samples of sodium hydrogen carbonate are heated until they decompose by 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂, and the sodium carbonate left behind is collected and weighed (M of NaHCO₃ = 84.0 g/mol; M of Na₂CO₃ = 106.0 g/mol). Run 1: 42.0 g of NaHCO₃, 7.685 g of Na₂CO₃ collected. Run 2: 16.8 g of NaHCO₃, 7.526 g of Na₂CO₃ collected. Which run has the higher percent yield?
Run 2 — 71% compared with 29%. KEY
Run 1 — collecting more grams of Na₂CO₃ (7.685 g against 7.526 g) proves the higher percent yield.
Run 1 — 345% compared with 141%.
The runs tie — the same decomposition always gives the same percent yield, whatever the sample size.
STO-030 · medium · application · seq 21 · authored (FRQ-1 form E: fresh taught-class precipitation (NH₄)₃PO₄ + 3LiCl → Li₃PO₄ + 3NH₄Cl with the measured species the 1-coefficient phosphate and the 3LiCl coefficient as the ratio lure; Li₃PO₄ is a genuinely insoluble lithium salt; exact product 0.0918 mol → 0.092 mol at 2 s.f.) · frq
0.34 L of 0.27 M ammonium phosphate solution is treated with excess lithium chloride by (NH₄)₃PO₄ + 3LiCl → Li₃PO₄ + 3NH₄Cl.
0.34 L of 0.27 M ammonium phosphate solution is treated with excess lithium chloride by (NH₄)₃PO₄ + 3LiCl → Li₃PO₄ + 3NH₄Cl.
(a) Calculate how many moles of (NH₄)₃PO₄ are dissolved in the 0.34 L of solution. Show your working.
(b) Use the mole ratio from the balanced equation to calculate how many moles of Li₃PO₄ form. Show your working. Give your answer in moles to 2 significant figures.
Part a. Calculate how many moles of (NH₄)₃PO₄ are dissolved in the 0.34 L of solution. Show your working.
Model: Values: V = 0.34 L of solution, M = 0.27 mol/L — M is the molarity of the ammonium phosphate solution, not a molar mass.
n = M × V
n = 0.27 × 0.34
n = 0.0918 mol of ammonium phosphate dissolved.
Part b. Use the mole ratio from the balanced equation to calculate how many moles of Li₃PO₄ form. Show your working. Give your answer in moles to 2 significant figures.
Model: From the balanced equation, 1 mol of lithium phosphate forms for every 1 mol of ammonium phosphate.
moles of lithium phosphate = 0.0918 × (1 ÷ 1)
moles of lithium phosphate = 0.0918 mol
To 2 significant figures: 0.092 mol of lithium phosphate. The product stays in moles.
Full model answer. (a) Values: V = 0.34 L of solution, M = 0.27 mol/L — M is the molarity of the ammonium phosphate solution, not a molar mass.
n = M × V
n = 0.27 × 0.34
n = 0.0918 mol of ammonium phosphate dissolved.
(b) From the balanced equation, 1 mol of lithium phosphate forms for every 1 mol of ammonium phosphate.
moles of lithium phosphate = 0.0918 × (1 ÷ 1)
moles of lithium phosphate = 0.0918 mol
To 2 significant figures: 0.092 mol of lithium phosphate. The product stays in moles.
STO-029 · hard · application · seq 22 · authored (FRQ-2 form E: indium burn, 4:2 ratio; UNSEEDED) · frq
A boat holding 57.4 g of indium is heated until the indium burns in excess O₂ by 4In + 3O₂ → 2In₂O₃. After cooling, 31.23 g of In₂O₃ is collected.
A boat holding 57.4 g of indium is heated until the indium burns in excess O₂ by 4In + 3O₂ → 2In₂O₃. After cooling, 31.23 g of In₂O₃ is collected.
(a) Calculate the theoretical yield of In₂O₃. Show every step of your working. Report the theoretical yield to 3 significant figures.
(b) Calculate the percent yield of In₂O₃. Give your answer as a percentage to 3 significant figures.
Molar masses (exact): In 114.8 g/mol, In₂O₃ 277.6 g/mol.
Write your answers in plain numbers and words; no special symbols are needed.
Part a. Calculate the theoretical yield of In₂O₃. Show every step of your working. Report the theoretical yield to 3 significant figures.
Model: Values: mass of indium = 57.4 g; molar mass of indium = 114.8 g/mol; molar mass of indium oxide = 277.6 g/mol.
Equation: n = m ÷ M
Substitute: n = 57.4 ÷ 114.8
Calculate: n = 0.500 mol of indium
Mole ratio from the balanced equation: 2 mol of indium oxide forms from 4 mol of indium, so n = 0.500 × (2 ÷ 4) = 0.250 mol of indium oxide
Equation: m = n × M
Substitute: m = 0.250 × 277.6
Calculate: theoretical yield = 69.40 g, reported as 69.4 g to 3 significant figures
Part b. Calculate the percent yield of In₂O₃. Give your answer as a percentage to 3 significant figures.
Molar masses (exact): In 114.8 g/mol, In₂O₃ 277.6 g/mol.
Write your answers in plain numbers and words; no special symbols are needed.
Model: Values: actual yield = 31.23 g; theoretical yield = 69.40 g.
Equation: percent yield = (actual yield ÷ theoretical yield) × 100
Substitute: percent yield = (31.23 ÷ 69.40) × 100
Calculate: percent yield = 45.0% (the rounded 69.4 g gives the same 45.0%)
Full model answer. (a) Values: mass of indium = 57.4 g; molar mass of indium = 114.8 g/mol; molar mass of indium oxide = 277.6 g/mol.
Equation: n = m ÷ M
Substitute: n = 57.4 ÷ 114.8
Calculate: n = 0.500 mol of indium
Mole ratio from the balanced equation: 2 mol of indium oxide forms from 4 mol of indium, so n = 0.500 × (2 ÷ 4) = 0.250 mol of indium oxide
Equation: m = n × M
Substitute: m = 0.250 × 277.6
Calculate: theoretical yield = 69.40 g, reported as 69.4 g to 3 significant figures
(b) Values: actual yield = 31.23 g; theoretical yield = 69.40 g.
Equation: percent yield = (actual yield ÷ theoretical yield) × 100
Substitute: percent yield = (31.23 ÷ 69.40) × 100
Calculate: percent yield = 45.0% (the rounded 69.4 g gives the same 45.0%)