OCR GCSE Chemistry A: Introducing chemical reactions (J248) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sub-topic C3.1 Introducing chemical reactions of the OCR GCSE (9-1) Chemistry A (Gateway Science) J248 specification, Version 4.0 (August 2026): statements C3.1a to C3.1l and the maths outcomes CM3.1i to CM3.1iv. Questions on C3.1e, C3.1g, C3.1h, C3.1k and C3.1l are marked (Higher Tier only); the others suit both tiers. Relative atomic masses are given in each question where you need them.

Before you start, read the C3.1 study guide or the C3.1 revision notes. The J248 course page and statement-by-statement checklist show where this sits, and the free diagnostics can point you to other weak areas.

Questions

1. Use the ions Li⁺, Mg²⁺, NH₄⁺, F⁻, OH⁻ and SO₄²⁻ to write the formula of each compound.

(a) lithium fluoride [1] (b) magnesium hydroxide [1] (c) ammonium sulfate [1]

2. Balance each equation.

(a) Na + O₂ → Na₂O [1] (b) C₄H₁₀ + O₂ → CO₂ + H₂O (the combustion of butane) [2]

3. Half equations.

(a) Write the half equation for zinc ions, Zn²⁺, forming zinc atoms. [1] (b) Write the half equation for iodide ions forming iodine molecules. [1] (c) Explain why there is a 2 in front of the iodide ion in your answer to (b). [1]

4. Aluminium reacts with dilute hydrochloric acid to give a solution of aluminium chloride, AlCl₃, and hydrogen gas. Write the balanced equation, including state symbols. [3]

5. (Higher Tier only)

(a) Iron(III) chloride solution is added to sodium hydroxide solution. A solid of iron(III) hydroxide, Fe(OH)₃, forms. Write the ionic equation, including state symbols. [2] (b) Chlorine solution is added to potassium bromide solution. The solution turns orange as bromine forms. Write the ionic equation. [2] (c) Name the spectator ion in reaction (b). [1]

6. A student puts zinc carbonate and dilute hydrochloric acid in a conical flask and seals it with a bung. The flask and contents have a mass of 85.62 g. Fizzing is seen until the reaction stops.

(a) State the mass of the sealed flask and contents at the end. [1] (b) Explain your answer to (a). [1] (c) The student removes the bung and weighs the flask again later. The reading is 85.18 g. Calculate the change in mass. [1] (d) Explain the change in (c) using the particle model. [2]

7. A crucible containing copper powder has a mass of 24.50 g. It is heated in air without a lid, then cooled and reweighed. The new mass is 24.82 g.

(a) Calculate the change in mass. [1] (b) Explain why the mass increased, using ideas about particles. [2]

8. (Higher Tier only) Use the Avogadro constant 6.02 × 10²³ per mole.

(a) State what is meant by the Avogadro constant. [1] (b) Calculate the mass of one silver atom. Ar of Ag = 108. Write the answer in standard form, to 3 significant figures. [2] (c) Calculate the number of atoms in 2.70 g of aluminium. Ar of Al = 27. [2]

9. (Higher Tier only) Use Ar: H = 1, O = 16, Mg = 24, S = 32, Ca = 40.

(a) Calculate the amount, in moles, of calcium hydroxide, Ca(OH)₂, in 2.96 g. [2] (b) Calculate the mass of 0.125 mol of magnesium sulfate, MgSO₄. [2]

10. (Higher Tier only) A student heats 0.84 g of lithium in nitrogen. The lithium reacts completely with 0.56 g of nitrogen, N₂, to make 1.40 g of lithium nitride, Li₃N. Use Ar: Li = 7, N = 14. Use the masses to deduce the balanced equation for the reaction. Show your working. [5]

11. (Higher Tier only) Aluminium reacts with iodine to form aluminium iodide, AlI₃. A teacher mixes 0.81 g of aluminium with 6.35 g of iodine, I₂. Use Ar: Al = 27, I = 127.

(a) Write the balanced equation for the reaction. [1] (b) Show that iodine is the limiting reactant. [3] (c) Calculate the maximum mass of aluminium iodide that can form. [2] (d) Calculate the mass of aluminium left over. [1] (e) The teacher repeats the reaction with 1.62 g of aluminium and the same mass of iodine. Explain why the mass of product does not change. [1]

12. (Higher Tier only) A strip of magnesium ribbon is left in silver nitrate solution, AgNO₃(aq). Grey crystals of silver grow on the strip, and magnesium nitrate, Mg(NO₃)₂, stays dissolved. Use Ar: N = 14, O = 16, Mg = 24, Ag = 108.

(a) Write the ionic equation, including state symbols. [2] (b) Calculate the mass of magnesium that reacts when 5.40 g of silver forms. [3] (c) Calculate the mass of magnesium nitrate made at the same time. [2]

Answers

1. (a) One Li⁺ balances one F⁻: LiF [1] (b) One Mg²⁺ needs two OH⁻, with brackets: Mg(OH)₂ [1] (c) Two NH₄⁺ balance one SO₄²⁻: (NH₄)₂SO₄ [1] [3] Examiner insight: A formula without the brackets, such as MgOH₂, usually gains no credit because it describes a different particle.

2. (a) 4Na + O₂ → 2Na₂O [1] (b) Carbon and hydrogen balanced: C₄H₁₀ + O₂ → 4CO₂ + 5H₂O, with oxygen still unbalanced [1]; doubling to clear the half: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O [1] [3] Examiner insight: Answers with 6½O₂ are correct in ratio, but whole-number coefficients are the expected form, so double if a half appears.

3. (a) Zn²⁺ + 2e⁻ → Zn [1] (b) 2I⁻ → I₂ + 2e⁻ [1] (c) Iodine exists as I₂ molecules, so two iodide ions are needed to make one molecule (and two electrons are released) [1] [3] Examiner insight: Check charges: a half equation that balances atoms but not charge, such as 2I⁻ → I₂ + e⁻, is not accepted as balanced.

4. Correct formulae: Al, HCl, AlCl₃ and H₂ [1]; balanced: 2Al + 6HCl → 2AlCl₃ + 3H₂ [1]; state symbols: 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g) [1] [3] Examiner insight: Get the formulae right before balancing; writing AlCl in place of the given AlCl₃ spoils the whole equation, however good the state symbols are.

5. (a) Correct species: Fe³⁺ and OH⁻ forming Fe(OH)₃ [1]; balanced with states: Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s) [1] (b) Correct species: Cl₂ and Br⁻ forming Cl⁻ and Br₂ [1]; balanced: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂ [1] (c) Potassium ion, K⁺ [1] [5] Examiner insight: An ionic equation that still contains Na⁺ or K⁺ on both sides is not fully simplified; remove every spectator ion.

6. (a) 85.62 g [1] (b) The flask is sealed, so no particles can enter or leave; total mass is conserved [1] (c) 85.62 − 85.18 = 0.44 g decrease [1] (d) Carbon dioxide gas formed in the reaction and its molecules escaped from the open flask into the air [1]; the atoms were not destroyed, so total mass is still conserved but the escaped gas is no longer weighed [1] [5] Examiner insight: Naming the gas and saying it left the container is what earns credit; “mass was lost in the reaction” contradicts conservation of mass.

7. (a) 24.82 − 24.50 = 0.32 g increase [1] (b) Oxygen molecules from the air react with the copper to form copper oxide [1]; the oxygen atoms are now bonded in the solid in the crucible, so their mass is added to the reading [1] [3] Examiner insight: Full credit needs both where the extra mass comes from (oxygen in the air) and why the balance now detects it.

8. (a) The number of particles (atoms, molecules or ions) in one mole of a substance, about 6.02 × 10²³ per mole [1] (b) Mass of one atom = 108 ÷ (6.02 × 10²³) [1] = 1.79 × 10⁻²² g [1] (c) Amount = 2.70 ÷ 27 = 0.100 mol [1]; atoms = 0.100 × 6.02 × 10²³ = 6.02 × 10²² atoms [1] [5] Examiner insight: A mass of 1.79 g, or a positive power of ten, shows the standard form went wrong; one atom must have a tiny mass.

9. (a) Mr of Ca(OH)₂ = 40 + 2 × (16 + 1) = 74 [1]; amount = 2.96 ÷ 74 = 0.0400 mol [1] (b) Mr of MgSO₄ = 24 + 32 + 4 × 16 = 120 [1]; mass = 0.125 × 120 = 15.0 g [1] [4] Examiner insight: An Mr of 57 for Ca(OH)₂ comes from ignoring the bracket; show the Mr working so a slip there is visible.

10. Amount of Li = 0.84 ÷ 7 = 0.12 mol [1]; amount of N₂ = 0.56 ÷ 28 = 0.020 mol [1]; Mr of Li₃N = 35, so amount = 1.40 ÷ 35 = 0.040 mol [1]; dividing by 0.020 gives the ratio Li : N₂ : Li₃N = 6 : 1 : 2 [1]; equation: 6Li + N₂ → 2Li₃N [1] [5] Examiner insight: Using 14 instead of 28 for nitrogen gas halves the ratio for N₂; always use the formula as it appears in the equation.

11. (a) 2Al + 3I₂ → 2AlI₃ [1] (b) Amount of Al = 0.81 ÷ 27 = 0.030 mol [1]; amount of I₂ = 6.35 ÷ 254 = 0.025 mol [1]; 0.030 mol Al would need 0.030 × 3/2 = 0.045 mol I₂, but only 0.025 mol is present, so iodine is limiting [1] (c) Amount of AlI₃ = 0.025 × 2/3 = 0.01667 mol; Mr of AlI₃ = 27 + 3 × 127 = 408 [1]; mass = 0.01667 × 408 = 6.80 g [1] (d) Al used = 0.01667 × 27 = 0.45 g, so Al left = 0.81 − 0.45 = 0.36 g [1] (e) Iodine is still the limiting reactant and its amount has not changed, so the same amount of product forms; the extra aluminium is simply left over [1] [8] Examiner insight: In “show that” parts, the comparison with the equation ratio must be written out; stating “iodine has fewer moles” without the 3 : 2 ratio does not prove it is limiting.

12. (a) Correct species, with nitrate ions left out: Mg, Ag⁺, Mg²⁺, Ag [1]; balanced with states: Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s) [1] (b) Amount of Ag = 5.40 ÷ 108 = 0.0500 mol [1]; ratio Mg : Ag = 1 : 2, so 0.0250 mol Mg [1]; mass = 0.0250 × 24 = 0.600 g [1] (c) Mr of Mg(NO₃)₂ = 24 + 2 × (14 + 3 × 16) = 148 [1]; 0.0250 mol × 148 = 3.70 g [1] [7] Examiner insight: Forgetting the 1 : 2 ratio and using 0.0500 mol of magnesium doubles both later answers; write the ratio line out explicitly.

Where marks are usually lost

• Writing formulae without brackets, such as NH₄₂SO₄ for ammonium sulfate.
• Changing subscripts to balance an equation instead of adding coefficients.
• Writing I or Cl instead of I₂ or Cl₂ in equations, and using 127 instead of 254 as the Mr of iodine.
• Half equations with electrons written as products when they should be reactants, or with unequal total charge.
• Leaving spectator ions in an ionic equation, or splitting a precipitate into ions.
• Explaining a mass change as “mass destroyed” rather than “gas escaped” or “oxygen gained from the air”.
• Comparing masses instead of moles to find the limiting reactant.
• Final answers with too many significant figures, or in ordinary form when the value is very small.

Next steps

• Recap the methods in the C3.1 revision notes.
• Re-read any weak section in the C3.1 study guide.
• See the rest of the course on the J248 course page.
• Tick off C3.1 on the J248 checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Source document: the J248 specification for GCSE (9-1) Chemistry A (Gateway Science) from OCR, Version 4.0, dated August 2026 and in use since first assessment in 2018. This page follows its sub-topic C3.1 Introducing chemical reactions.