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The Mole Concept and Stoichiometry

Topic 3 of 3

Reacting quantities

Identify limiting amounts and infer unknown relationships.

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

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 example

The 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.

  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.

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 understandingIn the worked example, would doubling the carbonate mass double the carbon dioxide produced?Think it through, then reveal the answer
No. HCl already limits the reaction. More carbonate would remain unreacted while the maximum CO2 amount stays 0.0150 mol, unless more acid is supplied.

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 example

Infer 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.

  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.

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 example

A 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.

  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.

Worked example

Deduce a stoichiometric ratio from experiment

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

  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.