The problems are on the reacting masses worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Reacting masses come from a balanced equation. These worked answers correct the 20 problems of the Chemistry 9-10 worksheet on reacting masses, from reading mole ratios to Lead iodide precipitate: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit. Compare your method as well as your final value: a correct method with a slip in arithmetic still earns most of the marks in an exam.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: HS-PS1-7, GCSE C3
Problem 1 Reading mole ratios
1. The coefficients give the ratio 1 : 3 : 2.2. n(H₂) = 4 × 3 = 12 mol.
3. n(NH₃) = 4 × 2 = 8 mol.
Problem 2 Relative formula masses
1. 40 + 12 + 3 × 16 = 100.2. 2 × 23 + 32 + 4 × 16 = 46 + 32 + 64 = 142.
3. 2 × 27 + 3 × 16 = 54 + 48 = 102.
Problem 3 Units of concentration
1. 250 ÷ 1000 = 0.250 dm³.2. c = n ÷ V = 0.50 ÷ 2.0 = 0.25 mol/dm³.
3. n = c × V = 0.80 × 0.250 = 0.20 mol.
Problem 4 A simple yield
1. It is the theoretical yield.2. 3.6 ÷ 4.5 × 100 = 80%.
3. 0.65 × 8.0 = 5.2 g.
Problem 5 Hydrogen from magnesium
1. n = 2.4 ÷ 24 = 0.10 mol.2. The ratio Mg : H₂ is 1 : 1, so 0.10 mol.
3. m = 0.10 × 2 = 0.20 g.
Problem 6 Mass of iron(III) oxide
1. 2 × 56 + 3 × 16 = 112 + 48 = 160 g/mol.2. m = 0.25 × 160 = 40 g.
3. n = 80 ÷ 160 = 0.50 mol.
Problem 7 Oxygen from potassium chlorate
1. 39 + 35.5 + 3 × 16 = 122.5 g/mol.2. n = 12.25 ÷ 122.5 = 0.100 mol.
3. n(O₂) = 0.100 × 3/2 = 0.150 mol, so m = 0.150 × 32 = 4.8 g.
Problem 8 Which gas runs out?
1. Hydrogen would need 6 ÷ 2 = 3 mol of O₂, and 4 mol is available, so hydrogen is limiting.2. n(H₂O) = n(H₂) = 6 mol.
3. O₂ used = 3 mol, so 1 mol of oxygen is left.
Problem 9 Burning methane
1. n = 8.0 ÷ 16 = 0.50 mol.2. n(CO₂) = 0.50 mol; M = 44 g/mol; m = 22 g.
3. n(H₂O) = 1.0 mol; M = 18 g/mol; m = 18 g.
Problem 10 Salt water for an aquarium
1. M = 58.5 g/mol, so n = 5.85 ÷ 58.5 = 0.100 mol.2. c = 0.100 ÷ 0.500 = 0.200 mol/dm³.
3. 5.85 ÷ 0.500 = 11.7 g/dm³ (or 0.200 × 58.5).
Problem 11 Weighing copper sulfate
1. n = 0.100 × 0.250 = 0.0250 mol.2. 63.5 + 32 + 64 = 159.5 g/mol.
3. m = 0.0250 × 159.5 = 3.99 g.
Problem 12 Silver chloride precipitate
1. M(AgNO₃) = 108 + 14 + 48 = 170, so n = 3.40 ÷ 170 = 0.0200 mol.2. n(AgCl) = 0.0200 mol; M = 143.5; m = 2.87 g.
3. 2.58 ÷ 2.87 × 100 = 89.9%.
Problem 13 Why is the yield low?
1. 0.72 × 15.0 = 10.8 g.2. Product is lost when transferring, filtering or evaporating; some reactant forms side products; the reaction may not go to completion because it is reversible.
3. More product than the equation allows means the solid is probably wet or impure, because mass cannot be created.
Problem 14 Which solution is strongest?
1. A: 0.04 mol ÷ 0.100 dm³ = 0.40 mol/dm³. B: 0.04 ÷ 0.050 = 0.80 mol/dm³. C: 0.06 ÷ 0.200 = 0.30 mol/dm³.2. B > A > C.
3. B must reach the concentration of A (0.40 mol/dm³), so its 0.04 mol needs 0.100 dm³ in total: add 50 cm³ of water.
Problem 15 Three trials, one mean
1. 3.85 ÷ 5.00 = 77.0%; 4.10 ÷ 5.00 = 82.0%; 3.95 ÷ 5.00 = 79.0%.2. (77.0 + 82.0 + 79.0) ÷ 3 = 79.3%.
3. Different losses when transferring and drying the solid, and small differences in the amounts weighed.
Problem 16 Precipitating chalk
1. n(Na₂CO₃) = 0.40 × 0.0250 = 0.010 mol; n(CaCl₂) = 0.30 × 0.0500 = 0.015 mol.2. The ratio is 1 : 1, so sodium carbonate (0.010 mol) is limiting.
3. n(CaCO₃) = 0.010 mol; M = 100; m = 1.0 g. CaCl₂ left = 0.015 − 0.010 = 0.005 mol.
Problem 17 Spot the error
1. He ignored the coefficient 2: each mole of CaCO₃ needs 2 mol of HCl.2. n(HCl) = 1.0 × 0.100 = 0.10 mol, which is the limiting amount. n(CO₂) = 0.10 ÷ 2 = 0.050 mol.
3. m(CO₂) = 0.050 × 44 = 2.2 g. Marble used = 0.050 mol × 100 = 5.0 g, so 5.0 g is left.
Problem 18 Roasting zinc ore
1. M(ZnS) = 97, n = 4.85 ÷ 97 = 0.0500 mol; n(ZnO) = 0.0500 mol; M(ZnO) = 81; theoretical = 4.05 g.2. 3.24 ÷ 4.05 × 100 = 80.0%.
3. n(O₂) = 0.0500 × 3/2 = 0.0750 mol, m = 2.40 g. n(SO₂) = 0.0500 mol; M = 64; m = 3.20 g. (Check: 4.85 + 2.40 = 7.25 = 4.05 + 3.20.)
Problem 19 Yield through three steps
1. Multiply the fractions: 0.90 × 0.80 × 0.75 = 0.540, so the overall yield is 54.0%.2. n(D) = 0.500 × 0.540 = 0.270 mol.
3. n(A) = 0.270 ÷ 0.540 = 0.500 mol.
Problem 20 Lead iodide precipitate
1. n(Pb(NO₃)₂) = 0.00500 mol; n(KI) = 0.00750 mol. KI would need to be 2 × 0.00500 = 0.0100 mol, so KI is limiting.2. n(PbI₂) = 0.00750 ÷ 2 = 0.00375 mol; M = 461; theoretical = 0.00375 × 461 = 1.73 g.
3. 1.40 ÷ 1.729 × 100 = 81.0%. Lead nitrate left = 0.00500 − 0.00375 = 0.00125 mol in 0.100 dm³, so 0.0125 mol/dm³.
