Reacting Mass Calculations Exam Questions
Practice IGCSE Chemistry exam questions on reacting mass calculations. Covers mass-to-mass calculations, limiting reagents, and using balanced equations to calculate masses of products and reactants with mark schemes and examiner notes.
Published by IGCSEChemistry.com.my
Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).
Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-19.
These questions cover all aspects of reacting mass calculations. Show all working clearly.
Question 1 (3 marks, Supplement)
Magnesium reacts with hydrochloric acid:
Mg + 2HCl → MgCl2 + H2
Calculate the mass of magnesium chloride produced when 4.8 g of magnesium reacts completely. (Ar: Mg = 24, Cl = 35.5)
Mark scheme
- Moles of Mg = 4.8 / 24 = 0.2 mol [1]
- From the equation, 1 mol Mg produces 1 mol MgCl2, so moles of MgCl2 = 0.2 mol [1]
- Mr of MgCl2 = 24 + (2 x 35.5) = 95; mass = 0.2 x 95 = 19.0 g [1]
Examiner note: The method is always: (1) find moles of given substance, (2) use the mole ratio from the balanced equation, (3) convert moles of required substance to mass. Write out each step separately.
Question 2 (4 marks, Supplement)
Calcium carbonate decomposes when heated:
CaCO3 → CaO + CO2
(a) Calculate the mass of calcium oxide (CaO) produced from 25.0 g of calcium carbonate. (3)
(b) Calculate the volume of carbon dioxide gas produced at RTP. (1)
(Ar: C = 12, O = 16, Ca = 40)
Mark scheme
(a)
- Mr of CaCO3 = 100; moles = 25.0 / 100 = 0.25 mol [1]
- Mole ratio CaCO3 : CaO = 1 : 1, so moles of CaO = 0.25 mol [1]
- Mr of CaO = 56; mass = 0.25 x 56 = 14.0 g [1]
(b) Moles of CO2 = 0.25; volume = 0.25 x 24 = 6.0 dm3 [1]
Examiner note: Part (b) requires you to convert moles to gas volume at RTP (24 dm3 per mole). This is a common extension to reacting mass questions. Always check whether the answer should be in g or dm3.
Question 3 (4 marks, Supplement)
Iron reacts with steam:
3Fe + 4H2O → Fe3O4 + 4H2
Calculate the mass of iron needed to produce 46.4 g of iron(II,III) oxide (Fe3O4). (Ar: Fe = 56, O = 16)
Mark scheme
- Mr of Fe3O4 = (3 x 56) + (4 x 16) = 168 + 64 = 232 [1]
- Moles of Fe3O4 = 46.4 / 232 = 0.2 mol [1]
- Mole ratio Fe : Fe3O4 = 3 : 1, so moles of Fe = 0.2 x 3 = 0.6 mol [1]
- Mass of Fe = 0.6 x 56 = 33.6 g [1]
Examiner note: This is a reverse calculation — working from product to reactant. The mole ratio is 3:1, so multiply moles of product by 3 to find moles of Fe. Many candidates forget to use the mole ratio and simply multiply by Mr.
Question 4 (4 marks, Supplement)
A student reacts 5.6 g of iron with excess dilute sulfuric acid:
Fe + H2SO4 → FeSO4 + H2
(a) Calculate the mass of hydrogen gas produced. (3)
(b) Calculate the volume of hydrogen gas produced at RTP. (1)
(Ar: H = 1, Fe = 56)
Mark scheme
(a) Moles of Fe = 5.6 / 56 = 0.1 mol [1] Mole ratio Fe : H2 = 1 : 1, so moles of H2 = 0.1 mol [1] Mass of H2 = 0.1 x 2 = 0.2 g [1]
(b) Volume = 0.1 x 24 = 2.4 dm3 [1]
Examiner note: “Excess” acid means the iron is the limiting reagent — all the iron reacts. The mole ratio is 1:1. Remember that hydrogen gas is H2, not H, so Mr = 2.
Question 5 (4 marks, Supplement)
2.40 g of magnesium is added to 100 cm3 of 1.0 mol/dm3 hydrochloric acid.
Mg + 2HCl → MgCl2 + H2
(a) Calculate the number of moles of magnesium and the number of moles of HCl. (2)
(b) Identify the limiting reagent and explain your choice. (1)
(c) Calculate the volume of hydrogen gas produced at RTP. (1)
(Ar: Mg = 24)
Mark scheme
(a) Moles of Mg = 2.40 / 24 = 0.10 mol [1] Moles of HCl = (100/1000) x 1.0 = 0.10 mol [1]
(b) HCl is the limiting reagent [1] — the equation requires 2 mol HCl per mol Mg, but we have equal moles of each. 0.10 mol Mg requires 0.20 mol HCl, but only 0.10 mol is available.
(c) 0.10 mol HCl produces 0.10/2 = 0.05 mol H2; volume = 0.05 x 24 = 1.2 dm3 [1]
Examiner note: To identify the limiting reagent, compare the actual mole ratio to the required ratio from the equation. Here, 0.10/0.10 = 1:1 actual ratio, but 1:2 is required. HCl runs out first. Use the limiting reagent to calculate the product.
Question 6 (3 marks, Supplement)
Aluminium reacts with iron(III) oxide in the thermite reaction:
2Al + Fe2O3 → Al2O3 + 2Fe
Calculate the mass of iron that can be obtained from 32.0 g of iron(III) oxide. (Ar: O = 16, Fe = 56)
Mark scheme
- Mr of Fe2O3 = (2 x 56) + (3 x 16) = 160; moles = 32.0/160 = 0.20 mol [1]
- Mole ratio Fe2O3 : Fe = 1 : 2, so moles of Fe = 0.40 mol [1]
- Mass of Fe = 0.40 x 56 = 22.4 g [1]
Examiner note: The mole ratio is 1:2 (one formula unit of Fe2O3 produces two atoms of Fe). Do not confuse the subscript 2 in Fe2O3 with the coefficient in the equation.
Question 7 (3 marks, Supplement)
A student burns 6.0 g of carbon in excess oxygen:
C + O2 → CO2
(a) Calculate the mass of carbon dioxide produced. (2)
(b) The student collects only 19.8 g of carbon dioxide. Calculate the percentage yield. (1)
(Ar: C = 12, O = 16)
Mark scheme
(a) Moles of C = 6.0/12 = 0.5 mol [1] Mr of CO2 = 44; mass = 0.5 x 44 = 22.0 g [1]
(b) Percentage yield = (19.8/22.0) x 100 = 90.0% [1]
Examiner note: Percentage yield = (actual yield / theoretical yield) x 100. The theoretical yield (22.0 g) is what the calculation predicts; the actual yield (19.8 g) is what was obtained experimentally. The yield is always 100% or less.
What to revise if you scored below 5
If the step-by-step method was unclear, memorise: (1) moles of given, (2) use ratio, (3) convert to required. If you struggled with limiting reagent (Question 5), practise comparing actual and required mole ratios. Revisit the reacting mass calculations notes.
Studying this yourself? Tutoring arrangements are normally made by a parent or guardian. Message us for the details to share with them, or send them this page.