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8873 / 2027
The Mole Concept and Stoichiometry

Lesson 3 of 3

Reacting quantities

Identify limiting amounts and infer unknown relationships.

A-Level 8873, revised syllabus (2026-2027)

  • I can calculate reacting quantities.Syllabus

    Syllabus 8873, 5(i). Calculate reacting quantities.

  • I can infer stoichiometric relationships.Syllabus

    Syllabus 8873, 5(j). Infer stoichiometric relationships.

Find the limiting reactant before the product

Compare amounts divided by stoichiometric coefficients.

One reliable calculation route
  1. Mass, solution or gas data

    Convert each relevant quantity into moles.

  2. Balanced equation

    Use coefficient ratios, not mass ratios.

  3. Limiting reactant

    Compare available moles divided by the coefficient for each reactant.

  4. Requested quantity

    Convert the product or excess amount into the requested units.

Worked exampleThe smaller mass is not automatically limiting

2.50 g CaCO3 is added to 50.0 cm3 of 0.600 mol dm-3 HCl. Find CO2 mass and unreacted CaCO3, using M(CaCO3) = 100.0 g mol-1.

Try it first. Then show the worked stepsHide the worked steps
  1. CaCO3 + 2HCl → CaCl2 + CO2 + H2O.
  2. CaCO3 amount = 2.50/100.0 = 0.0250 mol. HCl amount = 0.600 × 0.0500 = 0.0300 mol.
  3. HCl can react with only 0.0300/2 = 0.0150 mol CaCO3, so HCl is limiting.
  4. CO2 amount = 0.0150 mol; mass = 0.0150 × 44.0 = 0.660 g.
  5. CaCO3 remaining = 0.0250 - 0.0150 = 0.0100 mol, equivalent to 1.00 g.
Answer

0.660 g CO2 forms and 1.00 g CaCO3 remains. At an explicitly supplied molar gas volume of 24.0 dm3 mol-1, the gas volume would be 0.360 dm3.

Try it step by step

Work it out in steps: Moles of the carbonate, then Moles of acid, then Moles of CO2 from the limiting reactant, then Mass of CO2.

  1. Problem 1. 1.50 g of CaCO3 is added to 40.0 cm3 of 0.500 mol dm-3 HCl. CaCO3 + 2HCl -> CaCl2 + CO2 + H2O. Find the mass of CO2, in g to 3 significant figures. Use M(CaCO3) = 100.0 and M(CO2) = 44.0 g mol-1.
  2. Problem 2. 3.00 g of CaCO3 is added to 50.0 cm3 of 2.00 mol dm-3 HCl. Use the same equation. Find the mass of CO2, in g to 3 significant figures.
  3. Problem 3. 2.12 g of Na2CO3 is added to 30.0 cm3 of 1.00 mol dm-3 HCl. Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O. Use M(Na2CO3) = 106.0 and M(CO2) = 44.0 g mol-1. Give each answer to 3 significant figures.
Show the answer

Problem 1.

Moles of the carbonate: n(CaCO3) = 1.50 / 100.0 = 0.0150 mol.

Moles of acid: n(HCl) = 0.500 x 0.0400 = 0.0200 mol.

Moles of CO2 from the limiting reactant: HCl can react with only 0.0200 / 2 = 0.0100 mol CaCO3, so HCl is limiting and n(CO2) = 0.0100 mol.

Mass of CO2: Mass of CO2 = 0.0100 x 44.0 = 0.440 g.

Problem 2.

Moles of the carbonate: n(CaCO3) = 3.00 / 100.0 = 0.0300 mol.

Moles of acid: n(HCl) = 2.00 x 0.0500 = 0.100 mol.

Moles of CO2 from the limiting reactant: 0.0300 mol CaCO3 needs only 0.0600 mol HCl, so CaCO3 is limiting and n(CO2) = 0.0300 mol.

Mass of CO2: Mass of CO2 = 0.0300 x 44.0 = 1.32 g.

Problem 3.

Moles of the carbonate: n(Na2CO3) = 2.12 / 106.0 = 0.0200 mol.

Moles of acid: n(HCl) = 1.00 x 0.0300 = 0.0300 mol.

Moles of CO2 from the limiting reactant: HCl can react with only 0.0150 mol Na2CO3, so HCl is limiting and n(CO2) = 0.0300 / 2 = 0.0150 mol.

Mass of CO2: Mass of CO2 = 0.0150 x 44.0 = 0.660 g.

If the sample is impure, only the reacting component contributes its amount to the equation. If an isolated product yield is below the theoretical amount, percentage yield = actual amount/theoretical amount × 100%, using consistent units. Neither purity nor yield changes the stoichiometric coefficients.

Check your understanding

In the worked example, HCl was the limiting reactant. Would doubling the mass of carbonate double the CO2 produced?

  1. No. The acid still runs out, so the CO2 stays at 0.0150 mol.
  2. Yes. Twice the carbonate releases twice the carbon dioxide.
  3. No. It halves the CO2, because the acid is now more dilute.
Show the answer

No. The acid still runs out, so the CO2 stays at 0.0150 mol.

The amount of product is set by the limiting reactant. Extra carbonate would be left over unless more acid is added.

A gas volume can reveal an unknown formula

Compare volumes only at the same temperature and pressure.

For gases measured at the same temperature and pressure, volume is proportional to amount. Use a supplied molar gas volume Vm through n = V/Vm; keep cm3 and dm3 consistent. Do not silently assume a molar gas volume for unspecified conditions.

Complete combustion follows CxHy + (x + y/4)O2 → xCO2 + (y/2)H2O. If the products are cooled enough to condense water, that water no longer contributes to the measured gas volume. Carbon dioxide and unreacted oxygen still do.

Worked exampleInfer the hydrocarbon from gas data

20.0 cm3 of a gaseous hydrocarbon burns completely in 100.0 cm3 O2. After cooling to the original conditions and condensing water, 80.0 cm3 gas remains. Removing CO2 leaves 40.0 cm3 O2. Find the hydrocarbon formula.

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  1. CO2 formed = 80.0 - 40.0 = 40.0 cm3. The hydrocarbon:CO2 ratio is 20.0:40.0 = 1:2, so x = 2.
  2. O2 consumed = 100.0 - 40.0 = 60.0 cm3. The hydrocarbon:O2 ratio is 1:3.
  3. Thus x + y/4 = 3. With x = 2, y = 4.
  4. Check: C2H4 + 3O2 → 2CO2 + 2H2O. The final dry gas is 40.0 CO2 plus 40.0 excess O2.
Answer

C2H4. All volume comparisons assume the same temperature and pressure and complete combustion.

Try it step by step

Work it out in steps: Volume of CO2 formed, then Volume of O2 used, then Formula.

  1. Problem 1. 10.0 cm3 of a gaseous hydrocarbon CxHy burns completely in 60.0 cm3 of O2. After cooling to the starting conditions, 45.0 cm3 of gas remains. Removing CO2 leaves 15.0 cm3 of O2. Use CxHy + (x + y/4)O2 -> xCO2 + (y/2)H2O. Give volumes to 3 significant figures and x and y as whole numbers.
  2. Problem 2. 20.0 cm3 of a gaseous hydrocarbon burns completely in 150.0 cm3 of O2. After cooling, 110.0 cm3 of gas remains. Removing CO2 leaves 30.0 cm3 of O2. Find x and y as whole numbers, then the formula.
  3. Problem 3. 10.0 cm3 of a gaseous hydrocarbon burns completely in 80.0 cm3 of O2. After cooling, 60.0 cm3 of gas remains. Removing CO2 leaves 30.0 cm3 of O2. Give each volume in cm3 to 3 significant figures, then choose the formula.
Show the answer

Problem 1.

Volume of CO2 formed: CO2 = 45.0 - 15.0 = 30.0 cm3, so x = 30.0 / 10.0 = 3.

Volume of O2 used: O2 used = 60.0 - 15.0 = 45.0 cm3, so x + y/4 = 4.5 and y = 6.

Formula: The hydrocarbon is C3H6.

Problem 2.

Volume of CO2 formed: CO2 = 110.0 - 30.0 = 80.0 cm3, so x = 80.0 / 20.0 = 4.

Volume of O2 used: O2 used = 150.0 - 30.0 = 120.0 cm3, so x + y/4 = 6 and y = 8.

Formula: The hydrocarbon is C4H8.

Problem 3.

Volume of CO2 formed: CO2 = 60.0 - 30.0 = 30.0 cm3, so x = 3.

Volume of O2 used: O2 used = 80.0 - 30.0 = 50.0 cm3, so x + y/4 = 5 and y = 8.

Formula: The hydrocarbon is C3H8.

Make dilution and titration steps explicit

A pipette transfers an amount; a burette delivers the amount needed to react with it.

For a solution, n = cV with concentration in mol dm-3 and volume in dm3. Divide a volume in cm3 by 1000 first. Dilution changes concentration and volume but not the solute amount: c1V1 = c2V2 when no reaction or loss occurs.

Worked exampleA redox titration binds volume to a 1:5 ratio

25.0 cm3 Fe2+ solution requires 14.60 cm3 of 0.0200 mol dm-3 acidified MnO4-. Find the Fe2+ concentration.

Try it first. Then show the worked stepsHide the worked steps
  1. Permanganate amount = 0.0200 × 0.01460 = 2.92 × 10-4 mol.
  2. The balanced acidic equation requires five Fe2+ per MnO4-, so Fe2+ amount = 1.46 × 10-3 mol.
  3. That amount is in 0.0250 dm3. Concentration = 1.46 × 10-3 / 0.0250.
Answer

[Fe2+] = 0.0584 mol dm-3. The answer has three significant figures, matching the limiting input precision.

Try it step by step

Work it out in steps: Moles of MnO4-, then Moles of Fe2+ (5 per MnO4-), then Concentration of Fe2+.

  1. Problem 1. 25.0 cm3 of Fe2+ solution needs 20.00 cm3 of 0.0200 mol dm-3 acidified KMnO4. MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+. Find [Fe2+] in mol dm-3 to 3 significant figures.
  2. Problem 2. 25.0 cm3 of Fe2+ solution needs 16.40 cm3 of 0.0100 mol dm-3 acidified KMnO4. Use the same equation. Find [Fe2+] in mol dm-3 to 3 significant figures.
  3. Problem 3. 20.0 cm3 of Fe2+ solution needs 12.50 cm3 of 0.0200 mol dm-3 acidified KMnO4. Use the same equation. Give each answer to 3 significant figures.
Show the answer

Problem 1.

Moles of MnO4-: n(MnO4-) = 0.0200 x 0.02000 = 0.000400 mol.

Moles of Fe2+ (5 per MnO4-): n(Fe2+) = 5 x 0.000400 = 0.00200 mol.

Concentration of Fe2+: [Fe2+] = 0.00200 / 0.0250 = 0.0800 mol dm-3.

Problem 2.

Moles of MnO4-: n(MnO4-) = 0.0100 x 0.01640 = 0.000164 mol.

Moles of Fe2+ (5 per MnO4-): n(Fe2+) = 5 x 0.000164 = 0.000820 mol.

Concentration of Fe2+: [Fe2+] = 0.000820 / 0.0250 = 0.0328 mol dm-3.

Problem 3.

Moles of MnO4-: n(MnO4-) = 0.0200 x 0.01250 = 0.000250 mol.

Moles of Fe2+ (5 per MnO4-): n(Fe2+) = 5 x 0.000250 = 0.00125 mol.

Concentration of Fe2+: [Fe2+] = 0.00125 / 0.0200 = 0.0625 mol dm-3.

Worked exampleDeduce a stoichiometric ratio from experiment

0.0100 mol of an unknown acid reacts exactly with 0.0200 mol NaOH. What does the evidence suggest?

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  1. NaOH supplies one OH- per formula unit.
  2. Two moles of OH- react per mole of the acid.
  3. Under the stated complete neutralisation conditions, this is consistent with two replaceable protons per acid molecule.
Answer

The acid is dibasic in this neutralisation. The evidence gives a 1:2 ratio, not the acid's full molecular identity.

Use raw burette readings to obtain the delivered volume, and use concordant titres rather than an isolated overshoot. Record measured precision, retain guard digits during calculation, then round the final result appropriately. Units such as mol, mol dm-3, g and cm3 communicate different quantities; a correct number with the wrong unit is not a correct result.