IGCSE · Practice questions · Reacting masses, yield and purity

Mass to moles to mass.

Six original Cambridge-style questions on reacting masses, gas volumes, limiting reactants, percentage yield and percentage purity. Extended content.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Moles, ratio, back again.

Step 1: moles = mass ÷ M_r (or gas volume ÷ 24).
Step 2: use the numbers in the balanced equation; M_r never includes them.
Step 3: mass = moles × M_r (or volume = moles × 24).
Limiting reactant first, then calculate the product from it.

Calculation

[3 marks]

Iron is made in the blast furnace: Fe₂O₃ + 3CO → 2Fe + 3CO₂. Calculate the mass of iron that can be made from 320 g of iron(III) oxide. (A_r: O 16, Fe 56)

M_r(Fe₂O₃) = 160, so moles = 320 ÷ 160 = 2 mol. ✓
Ratio Fe₂O₃ : Fe = 1 : 2, so 4 mol of Fe. ✓
Mass of Fe = 4 × 56 = 224 g. ✓

Calculation

[3 marks]

Calculate the volume of carbon dioxide, at r.t.p., given off when 25 g of calcium carbonate is heated until it has all decomposed. CaCO₃ → CaO + CO₂ (A_r: C 12, O 16, Ca 40)

M_r(CaCO₃) = 100, so moles = 25 ÷ 100 = 0.25 mol. ✓
Ratio 1 : 1, so 0.25 mol of CO₂. ✓
Volume = 0.25 × 24 = 6 dm³. ✓

Analysis

[4 marks]

6.5 g of zinc is added to 3.65 g of hydrogen chloride dissolved in water. Zn + 2HCl → ZnCl₂ + H₂. (a) Show which reactant is limiting. (b) Calculate the volume of hydrogen formed at r.t.p. (c) Calculate the mass of the reactant left over. (A_r: H 1, Cl 35.5, Zn 65)

Zn: 6.5 ÷ 65 = 0.1 mol. HCl (M_r 36.5): 3.65 ÷ 36.5 = 0.1 mol. ✓
0.1 mol of Zn needs 0.2 mol of HCl, but only 0.1 mol is present, so HCl is limiting. ✓
0.1 mol HCl gives 0.05 mol H₂ = 0.05 × 24 = 1.2 dm³. ✓
0.05 mol of Zn reacts, so 0.05 mol is left: 0.05 × 65 = 3.25 g of zinc. ✓

Calculation

[2 marks]

A student burns 4.8 g of magnesium and expects to make 8.0 g of magnesium oxide. She collects 7.2 g. Calculate the percentage yield.

Percentage yield = actual ÷ expected × 100. ✓
7.2 ÷ 8.0 × 100 = 90%. ✓

Calculation

[3 marks]

10.0 g of impure calcium carbonate reacts with excess hydrochloric acid and gives 2.16 dm³ of carbon dioxide at r.t.p. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Calculate the percentage purity of the calcium carbonate. (A_r: C 12, O 16, Ca 40)

Moles of CO₂ = 2.16 ÷ 24 = 0.09 mol. ✓
Ratio 1 : 1, so 0.09 mol of CaCO₃ = 0.09 × 100 = 9.0 g. ✓
Purity = 9.0 ÷ 10.0 × 100 = 90%. ✓

Explain

[2 marks]

Give two reasons why the actual yield of a product is usually less than the expected yield.

The reaction may not go to completion (for example, it is reversible) or other side reactions make different products. ✓
Some product is lost when it is separated, filtered, transferred or dried. ✓

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