The coefficients of a balanced chemical equation also fix how many moles of each substance react and form.

Working out reacting amounts from a balanced chemical equation is called 'stoichiometry'.

The 'mole ratio' of two substances is the ratio of their coefficients, taken in the order the two substances are named.

A substance written with no coefficient has the coefficient 1.

In 2H₂O → 2H₂ + O₂, the mole ratio of H₂ to O₂ is 2:1 — the demo's two test tubes to one.

In N₂ + 3H₂ → 2NH₃, the mole ratio of H₂ to NH₃ is 3:2.

Order matters: the mole ratio of NH₃ to H₂ is 2:3.

Subscripts play no part — the mole ratio comes from the coefficients alone.

Take a recipe ratio you can count: 2 eggs for every 1 cup of flour.

Scale both up by the same bundle — 2 dozen eggs for every 1 dozen cups of flour — and the two-to-one ratio is untouched.

A balanced equation is a recipe in particles: 2H₂ + O₂ → 2H₂O says 2 molecules of H₂ react for every 1 molecule of O₂.

A mole is the same fixed count of particles for every substance.

So switching the equation's counting unit from molecules to moles scales every substance up by that same count — 2 mol of H₂ react for every 1 mol of O₂.

The coefficient ratio applies to moles just as it applies to particles, because a mole is the same fixed count for every substance, so scaling every substance up by that same count leaves the ratio unchanged.

This is why the mole ratio read from coefficients is trustworthy for real, weighable amounts.

A mole ratio is only as good as the coefficients it comes from.

An unbalanced equation miscounts atoms — it shows atoms appearing or disappearing, and a reaction only rearranges atoms.

So an unbalanced equation has wrong coefficients, and wrong coefficients give a wrong mole ratio.

The unbalanced equation H₂ + O₂ → H₂O suggests 1 mol of O₂ makes 1 mol of H₂O.

The balanced equation 2H₂ + O₂ → 2H₂O shows the truth: 1 mol of O₂ makes 2 mol of H₂O.

The unbalanced prediction is only half the real amount — enough to ruin anything calculated from it.

So before reading any mole ratio, check that the equation is balanced.

The 'given substance' is the one whose moles you know; the 'wanted substance' is the one whose moles you are after.

The equation is: moles of wanted substance = moles of given substance × (coefficient of wanted substance ÷ coefficient of given substance).

The wanted substance's coefficient sits on top of the fraction; the given substance's coefficient sits on the bottom.

For N₂ + 3H₂ → 2NH₃, starting from moles of H₂ and wanting moles of NH₃: moles of NH₃ = moles of H₂ × (2 ÷ 3).

One equation covers every pair of substances in any balanced equation — only the two coefficients change.

Write down the moles of the given substance from the question.

Write down the mole-ratio equation with the two substances named.

Read the two coefficients from the balanced equation — wanted on top, given on the bottom.

Substitute and calculate — the result carries the unit mol.

In N₂ + 3H₂ → 2NH₃, 6.0 mol of H₂ makes: moles of NH₃ = 6.0 × (2 ÷ 3) = 4.0 mol.

The same equation answers the reverse question — how much reactant a target amount of product needs — with the wanted and given roles swapped.

Sanity check: when the wanted coefficient is smaller than the given one, the answer comes out smaller than the given moles.

A 'mass-to-mass' calculation finds the mass of one substance from the mass of another in the same reaction.

It runs in three steps, always in the same order.

Step 1: convert the given mass to moles, using n = m/M.

Step 2: convert moles of the given substance to moles of the wanted substance, using the mole-ratio equation.

Step 3: convert the wanted substance's moles to mass, using m = n × M.

For 'what mass of CO₂ does 8.0 g of CH₄ produce?', the route is: grams of CH₄ → moles of CH₄ → moles of CO₂ → grams of CO₂.

Only the middle step uses the balanced equation — each outer step uses that substance's own molar mass.

Step 1: convert the given mass to moles with n = m/M.

Step 2: convert those moles to moles of the wanted substance with the mole-ratio equation.

In CH₄ + 2O₂ → CO₂ + 2H₂O, 32.0 g of O₂ (M = 32.0 g/mol) is: n = 32.0 / 32.0 = 1.0 mol of O₂.

Then: moles of H₂O = 1.0 × (2 ÷ 2) = 1.0 mol.

Stop at moles — the question asks for moles, so there is no third step.

Label every mole value with its substance, so step 2 starts from the right one.

Step 1: convert the given moles to moles of the wanted substance with the mole-ratio equation.

Step 2: convert those moles to mass with m = n × M, the rearranged mole equation you've already seen.

In N₂ + 3H₂ → 2NH₃, 3.0 mol of H₂ makes: moles of NH₃ = 3.0 × (2 ÷ 3) = 2.0 mol.

Then: m of NH₃ = 2.0 × 17.0 = 34.0 g (M of NH₃ = 17.0 g/mol).

Use the WANTED substance's molar mass in step 2 — the given substance's molar mass never appears.

Sanity check: the moles crossing into step 2 must belong to the wanted substance, so label every mole value.

Run the three steps in order: mass to moles, mole ratio, moles to mass.

Label every intermediate value with its substance — mixing up whose moles they are is the classic slip.

In CH₄ + 2O₂ → CO₂ + 2H₂O, 8.0 g of CH₄ (M = 16.0 g/mol) is: n = 8.0 / 16.0 = 0.50 mol of CH₄.

Then: moles of CO₂ = 0.50 × (1 ÷ 1) = 0.50 mol.

Then: m of CO₂ = 0.50 × 44.0 = 22.0 g (M of CO₂ = 44.0 g/mol).

The final line carries the unit g — the route ends in grams because the question asked for grams.

The three steps are the same — only the starting point changes.

The given substance is now the product, because its mass is the one you know.

Step 1: convert the product's target mass to moles with n = m/M.

Step 2: convert to moles of the reactant with the mole-ratio equation.

Step 3: convert the reactant's moles to mass with m = n × M.

In 2Mg + O₂ → 2MgO, a target of 40.3 g of MgO (M = 40.3 g/mol) is: n = 40.3 / 40.3 = 1.0 mol of MgO.

Then: moles of Mg = 1.0 × (2 ÷ 2) = 1.0 mol.

Then: m of Mg = 1.0 × 24.3 = 24.3 g (M of Mg = 24.3 g/mol) — burn 24.3 g of magnesium to make 40.3 g of MgO.

A balanced chemical equation shows the mass of every substance it uses and makes.

The mass of each substance is its coefficient multiplied by its molar mass.

Add the masses on the reactant side to get the total mass of reactants.

Add the masses on the product side to get the total mass of products.

The two totals always come out equal — the balanced equation obeys the law of conservation of mass.

For CH₄ + 2O₂ → CO₂ + 2H₂O, the reactant side carries 16.0 + 2 × 32.0 = 80.0 g.

The product side carries 44.0 + 2 × 18.0 = 80.0 g — the two totals match.

If your two totals do not match, a coefficient or a molar mass slipped — recheck them.

A product's mass can be greater than the mass of the single reactant it was calculated from.

The extra mass comes from the other reactant.

Every atom in the product came from one of the reactants, because the atoms themselves are not created, destroyed, or changed — they only separate and regroup into new combinations.

In the burning strip, each magnesium atom joins with an oxygen atom from the air to make MgO.

Burning 24.3 g of magnesium uses 16.0 g of oxygen, and 24.3 + 16.0 = 40.3 g of MgO forms.

Counting the oxygen, no mass appeared from nowhere — the ash outweighs the metal by exactly the oxygen's mass.

A worked mass-to-mass calculation is checked one step at a time.

Check step 1: the moles must come from n = m/M, dividing the given mass by the given substance's own molar mass.

Check step 2: the mole ratio must be the coefficient of the wanted substance divided by the coefficient of the given substance, read from the balanced equation.

Check step 3: the mass must come from m = n × M, using the wanted substance's molar mass.

The most common error is a flipped mole ratio — 2/1 written where the balanced equation requires 1/2.

A wrong molar mass is the quieter error: the arithmetic still looks tidy, so check WHICH substance's molar mass each step uses.

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.

A reaction stops forming product when one reactant is completely used up.

It stops even if plenty of the other reactant remains.

Magnesium burning in a sealed jar shows this: the flame dies as the jar's O₂ runs out.

The leftover magnesium stays in the jar, unreacted — with no O₂ left to react with, it cannot burn.

No new product forms after that moment.

One honest detail: a real flame gutters out just before the very last trace of oxygen is used — a flame needs a steady supply — but the story is the same: the fixed oxygen supply ran out, so the burning stopped.

The reactant that is completely used up first — and so stops the reaction — is called the 'limiting reactant'.

The reactant that is left over when the reaction stops is called the 'excess reactant'.

When magnesium burns in a sealed jar until the O₂ is gone, the O₂ is the limiting reactant.

The magnesium left sitting in that jar is the excess reactant.

Every reaction that stops this way has exactly one of each role: one reactant runs out, the other is left over.

A particle diagram of a reactant mixture shows exactly how many particles of each reactant are present.

Count the particles of each reactant in the diagram.

The figure shows 6 H₂ molecules and 2 O₂ molecules for the reaction 2H₂ + O₂ → 2H₂O.

Use the coefficients to find how many particles of one reactant would be needed to use up the other.

Using up all 6 H₂ molecules would need 3 O₂ molecules, because the equation uses 2 H₂ for every 1 O₂.

Only 2 O₂ molecules are shown — fewer than the 3 needed.

So the O₂ runs out first: O₂ is the limiting reactant, and H₂ is the excess reactant.

The after-reaction diagram shows the product formed plus the leftover excess reactant — and none of the limiting reactant.

First identify the limiting reactant from the counts, as you've already practiced.

Then use the coefficients to find how many product particles the limiting reactant makes.

Then find how many particles of the excess reactant are used, and subtract to find how many remain.

The figure runs the whole routine for 6 H₂ and 2 O₂ reacting by 2H₂ + O₂ → 2H₂O.

The 2 O₂ are limiting; they make 4 H₂O and use 4 H₂, so 6 − 4 = 2 H₂ are left over.

The after box holds 4 H₂O and 2 H₂ — and no O₂, because the limiting reactant is completely used up.

The figure shows a reactant mixture for the reaction 2Mg + O₂ → 2MgO. On paper, draw the particle diagram of the mixture after the reaction stops, then select Continue to compare your drawing with the model answer.

Every particle of product is built from particles of every reactant in the equation.

The limiting reactant caps the product because product can only form while every reactant is still available, so the reactant that runs out first stops product formation.

Once the limiting reactant is gone, the leftover excess reactant has nothing to pair with.

In the sealed jar, once the O₂ is gone, no more MgO can form — no matter how much magnesium remains.

Adding more of the excess reactant changes nothing; only more of the limiting reactant would let more product form.

The reactant present in the smaller amount is not always the limiting reactant.

The coefficients decide how much of each reactant is needed.

A reactant that the equation demands in large ratio can run out first even when there is more of it.

Take 5 mol of H₂ with 2 mol of N₂ in N₂ + 3H₂ → 2NH₃.

Reacting all the N₂ would need 6 mol of H₂, and only 5 mol is available.

So H₂ is the limiting reactant — even though there is more H₂ than N₂.

Compare amounts only after scaling by the coefficients, never straight off.

Pick one reactant and calculate how much of the other reactant would be needed to use it up completely.

The mole-ratio equation you've already seen does the work: moles of wanted substance = moles of given substance × (coefficient of wanted substance ÷ coefficient of given substance).

Compare the moles needed with the moles available.

If less is available than needed, the other reactant runs out first — it is the limiting reactant.

If more is available than needed, the reactant you picked first is the limiting one, and the other is in excess.

For 2 mol of N₂ and 3 mol of H₂ in N₂ + 3H₂ → 2NH₃: using up 2 mol of N₂ would need 2 × 3/1 = 6 mol of H₂.

Only 3 mol of H₂ is available — less than the 6 mol needed — so H₂ is the limiting reactant.

The limiting reactant sets the amount of product, because product can only form while every reactant is still available, so the reactant that runs out first stops product formation.

So a product calculation starts from the moles of the limiting reactant — never from the excess reactant.

The mole-ratio equation you've already seen does the rest: moles of wanted substance = moles of given substance × (coefficient of wanted substance ÷ coefficient of given substance).

Here the given substance is the limiting reactant, and the wanted substance is the product.

If a question also states how much excess reactant is present, that amount does not enter the calculation.

In N₂ + 3H₂ → 2NH₃, suppose H₂ is the limiting reactant and 3.0 mol of H₂ is available.

Moles of NH₃ = 3.0 × (2 ÷ 3) = 2.0 mol — the most NH₃ this mixture can make.

When the masses of both reactants are given, first convert each mass to moles with n = m/M.

Identify the limiting reactant the way you've already practiced: use the mole ratio to find how much of one reactant is needed to use up the other, and compare with the amount available.

Calculate the moles of product from the moles of the limiting reactant with the mole-ratio equation.

Convert the product moles to mass with m = n × M.

Four steps, no new equations — every link is one you've already used.

Try it on 24.3 g of Mg and 24.0 g of O₂ reacting by 2Mg + O₂ → 2MgO (M of Mg = 24.3 g/mol; M of O₂ = 32.0 g/mol; M of MgO = 40.3 g/mol).

Step 1: n(Mg) = 24.3 ÷ 24.3 = 1.00 mol, and n(O₂) = 24.0 ÷ 32.0 = 0.750 mol.

Step 2: using up 1.00 mol of Mg needs 1.00 × (1 ÷ 2) = 0.500 mol of O₂, and 0.750 mol is available — so Mg is limiting.

Step 3: moles of MgO = 1.00 × (2 ÷ 2) = 1.00 mol.

Step 4: m(MgO) = 1.00 × 40.3 = 40.3 g.

The amount of product a stoichiometry calculation predicts is called the 'theoretical yield'.

For the methane run, the calculated 22 g of CO₂ is the theoretical yield.

The amount of product actually measured after the experiment is called the 'actual yield'.

For the methane run, the 11 g the balance reads is the actual yield.

The theoretical yield comes from a calculation; the actual yield comes from a balance.

The two amounts answer different questions: what the reaction could make, and what the experiment did make.

Between the reaction flask and the final weighing, product goes missing in three main ways.

Some product is lost on transfer — a little stays behind on every filter paper, beaker wall, and spatula.

Some reactions stop before all of the reactants have reacted, so less product ever forms.

Some reactant is used up by side reactions, which turn it into other substances instead of the wanted product.

Every one of these effects removes product or prevents it from forming — none of them adds any.

That is why the actual yield of a reaction is usually less than the theoretical yield.

The 'percent yield' is the actual yield divided by the theoretical yield, times 100.

percent yield = (actual yield ÷ theoretical yield) × 100

The actual yield goes on top — the measured amount is compared against the predicted amount.

Both yields are masses of the same product in the same unit, so the units cancel, and percent yield is a plain percentage.

An actual yield of 15 g with a theoretical yield of 20 g gives (15 ÷ 20) × 100 = 75%.

A percent yield of 100% would mean the experiment collected every gram the calculation predicted.

Write down the actual yield and the theoretical yield from the question.

Write down the equation: percent yield = (actual yield ÷ theoretical yield) × 100.

Substitute the values and calculate — the result carries the unit %.

An actual yield of 12 g with a theoretical yield of 48 g gives (12 ÷ 48) × 100 = 25%.

Sanity check: the actual yield is less than the theoretical yield, so the percent yield must come out below 100% — an answer above 100% means the division went the wrong way.

The percent yield equation you've already seen still applies: percent yield = (actual yield ÷ theoretical yield) × 100.

The question gives the percent yield and the theoretical yield and asks for the actual yield, so rearrange the equation.

Make the actual yield the subject: actual yield = (percent yield ÷ 100) × theoretical yield.

Then substitute and calculate — the result carries the unit g.

It is one equation, rearranged when needed — not a second equation to memorize.

A theoretical yield of 40 g at a 90% yield gives actual yield = (90 ÷ 100) × 40 = 36 g.

Sanity check: a percent yield below 100% must give an actual yield below the theoretical yield.

The theoretical yield is the most product the reactants can possibly make.

So the actual yield can never genuinely rise above the theoretical yield.

That caps a genuine percent yield at 100%.

A reported percent yield over 100% therefore signals a measurement or procedure error.

A common culprit: product weighed while still wet, so the balance reads product plus water.

A yield at or below 100% may be genuine; a yield above 100% never is.

The percent yield equation needs a theoretical yield, and the question only gives the starting mass of a reactant — so calculate the theoretical yield first.

Step 1: convert the reactant mass to moles with n = m/M.

Step 2: convert to moles of product with the mole ratio.

Step 3: convert the product moles to mass with m = n × M — that mass is the theoretical yield.

Step 4: divide the measured actual yield by the theoretical yield and multiply by 100.

The first three steps are exactly the mass-to-mass routine you've already mastered — the percent yield equation simply takes its answer.

For the methane run (8.0 g of CH₄ burned in excess O₂ by CH₄ + 2O₂ → CO₂ + 2H₂O, 11 g of CO₂ collected): the theoretical yield works out to 22 g, so percent yield = (11 ÷ 22) × 100 = 50%.

You've already seen how a label's molarity and a measured volume give moles: n = M × V.

In n = M × V, M stands for the molarity in mol/L — not the molar mass — and V is the volume of solution in liters.

Those moles feed the same mole-ratio equation as every other calculation in this unit.

Nothing new happens after the first step: moles are moles, whether they came from a balance or from a bottle.

For 0.50 L of 2.0 M HCl reacting with excess zinc by Zn + 2HCl → ZnCl₂ + H₂: n(HCl) = 2.0 × 0.50 = 1.0 mol.

Moles of H₂ = 1.0 × (1 ÷ 2) = 0.50 mol.