AS · Practice questions · Reacting masses and limiting reagents

Mass to moles to mass.

Six original Cambridge-style questions on reacting masses: the three-step route, limiting and excess reagents, percentage yield, deducing a mole ratio and working in tonnes.

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

Keep these straight

Moles in, ratio, moles out.

Mass → moles: n = m ÷ M.
Ratio: from the balanced equation, never from masses.
Limiting reagent: smallest (moles ÷ coefficient).
% yield = actual ÷ theoretical (from the limiting reagent) × 100.

Calculation

[2 marks]

Calculate the mass of carbon dioxide formed when 5.00 g of calcium carbonate reacts with excess hydrochloric acid. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. (M_r: CaCO₃ 100.1, CO₂ 44.0)

n(CaCO₃) = 5.00 ÷ 100.1 = 0.0500 mol, so n(CO₂) = 0.0500 mol. ✓
Mass of CO₂ = 0.0500 × 44.0 = 2.20 g. ✓

Calculation

[4 marks]

5.40 g of aluminium is heated with 14.2 g of chlorine. 2Al + 3Cl₂ → 2AlCl₃. Identify the limiting reagent and calculate the mass of aluminium chloride formed and the mass of the reagent left over. (A_r: Al 27.0, Cl 35.5)

n(Al) = 5.40 ÷ 27.0 = 0.200 mol; n(Cl₂) = 14.2 ÷ 71.0 = 0.200 mol. ✓
Dividing by the coefficients: Al 0.200 ÷ 2 = 0.100; Cl₂ 0.200 ÷ 3 = 0.0667. Chlorine is limiting. ✓
n(AlCl₃) = 0.200 × 2/3 = 0.133 mol; M_r = 133.5, so mass = 17.8 g. ✓
Al used = 0.133 mol, so 0.0667 mol is left: 0.0667 × 27.0 = 1.80 g of aluminium in excess. ✓

Calculation

[2 marks]

12.0 g of calcium carbonate is heated and 5.60 g of calcium oxide is collected. Calculate the percentage yield. (M_r: CaCO₃ 100.1, CaO 56.1)

Theoretical: n = 12.0 ÷ 100.1 = 0.120 mol, so 0.120 × 56.1 = 6.73 g of CaO. ✓
Percentage yield = 5.60 ÷ 6.73 × 100 = 83.3%. ✓

Analysis

[3 marks]

1.12 g of iron reacts completely with 2.13 g of chlorine, Cl₂, to form a single chloride. Use the mole ratio to deduce the balanced equation. (A_r: Cl 35.5, Fe 55.8)

n(Fe) = 1.12 ÷ 55.8 = 0.0201 mol; n(Cl₂) = 2.13 ÷ 71.0 = 0.0300 mol. ✓
Fe : Cl₂ = 0.0201 : 0.0300 = 2 : 3. ✓
2Fe + 3Cl₂ → 2FeCl₃. ✓

Calculation

[3 marks]

An iron ore contains 80.0% Fe₂O₃ by mass. Calculate the maximum mass of iron, in tonnes, that can be extracted from 1.00 tonne of the ore. Fe₂O₃ + 3CO → 2Fe + 3CO₂. (M_r: Fe₂O₃ 159.6; A_r: Fe 55.8; 1 tonne = 10⁶ g)

Mass of Fe₂O₃ = 0.800 × 10⁶ g, so n = 0.800 × 10⁶ ÷ 159.6 = 5.01 × 10³ mol. ✓
n(Fe) = 2 × 5.01 × 10³ = 1.00 × 10⁴ mol. ✓
Mass = 1.00 × 10⁴ × 55.8 = 5.59 × 10⁵ g = 0.559 tonnes. ✓

Explain

[3 marks]

Suggest three reasons why the actual yield of a product in a laboratory preparation is less than the theoretical yield.

The reaction may not go to completion, or may be reversible and reach equilibrium. ✓
Side reactions may convert some of the reactant into other products. ✓
Some product is lost during transfer, filtration, recrystallisation or other purification steps. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.