Limiting reactants (HT only) - AQA GCSE Chemistry Exam Questions

Practise 19 exam-style questions on Limiting reactants (HT only) for AQA GCSE Chemistry. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

19 questions
1–6 marks each
Instant marking
No sign-up needed

19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

In a chemical reaction involving two reactants, it is common to use an excess of one of the reactants.

Explain why.

Model answer

Using an excess of one reactant makes sure that all of the other reactant is used up. This means the maximum possible amount of product is made from it and none of it is wasted, which matters if it is the more expensive reactant.

Explanation

The reactant that is completely used up is the limiting reactant. Chemists usually choose the cheaper or more easily removed substance to be in excess, so that the valuable reactant is fully converted to product.

Mark scheme

1 mark
to make sure that all of the other reactant is used up / reacts
1 mark
so the maximum amount of product is obtained from it
allow none of the other (more expensive / valuable) reactant is wasted

Question 2

3 marks

Magnesium burns in oxygen:

2Mg + O₂ → 2MgO

A student heats 2.4 g of magnesium with 3.2 g of oxygen. Identify the limiting reactant and calculate the mass of magnesium oxide formed.

Relative atomic masses (Ar): O = 16, Mg = 24

Model answer

moles of Mg = 2.4 ÷ 24 = 0.10 mol
moles of O₂ = 3.2 ÷ 32 = 0.10 mol
The equation shows 2 mol Mg reacts with 1 mol O₂, so 0.10 mol Mg needs only 0.05 mol O₂. Oxygen is in excess and magnesium is the limiting reactant.
moles of MgO = moles of Mg = 0.10 mol
mass of MgO = 0.10 × 40 = 4.0 g

Explanation

Convert both masses to moles, then use the equation ratio to see which reactant runs out first. The amount of product is always calculated from the limiting reactant, never from the one in excess.

Mark scheme

1 mark
moles of Mg = 2.4 ÷ 24 = 0.10 and moles of O₂ = 3.2 ÷ 32 = 0.10
correct answer with no working scores full marks
1 mark
magnesium is limiting because 0.10 mol Mg needs only 0.05 mol O₂
allow oxygen is in excess
1 mark
mass of MgO = 0.10 × 40 = 4.0 g
allow ecf from the limiting reactant chosen

Question 3

3 marks

Magnesium reacts with hydrochloric acid:

Mg + 2HCl → MgCl₂ + H₂

4.8 g of magnesium is added to an acid solution containing 0.20 mol of HCl. Calculate the mass of magnesium left unreacted.

Relative atomic mass (Ar): Mg = 24

Model answer

moles of Mg = 4.8 ÷ 24 = 0.20 mol
The equation shows 2 mol HCl reacts with 1 mol Mg, so 0.20 mol HCl reacts with 0.10 mol Mg.
magnesium left = 0.20 - 0.10 = 0.10 mol
mass left = 0.10 × 24 = 2.4 g

Explanation

The acid is the limiting reactant here. To find how much excess reactant is left, work out how much of it reacts with the limiting reactant and subtract that from the starting amount.

Mark scheme

1 mark
moles of Mg = 4.8 ÷ 24 = 0.20 (mol)
1 mark
0.20 mol HCl reacts with 0.10 mol Mg, so 0.10 mol Mg is left
allow ecf
1 mark
mass left = 0.10 × 24 = 2.4 g
allow ecf; correct answer with no working scores 3 marks

Question 4

4 marks

A factory makes ammonium sulfate fertiliser:

2NH₃ + H₂SO₄ → (NH₄)₂SO₄

In one batch, 3.4 tonnes of ammonia react with 14.7 tonnes of sulfuric acid. Identify the limiting reactant and calculate the maximum mass of ammonium sulfate produced, in tonnes.

Relative formula masses (Mr): NH₃ = 17, H₂SO₄ = 98, (NH₄)₂SO₄ = 132
1 tonne = 1 000 000 g

Model answer

moles of NH₃ = 3 400 000 ÷ 17 = 200 000 mol
moles of H₂SO₄ = 14 700 000 ÷ 98 = 150 000 mol
200 000 mol NH₃ needs only 100 000 mol H₂SO₄, so ammonia is the limiting reactant and sulfuric acid is in excess.
moles of (NH₄)₂SO₄ = 200 000 ÷ 2 = 100 000 mol
mass = 100 000 × 132 = 13 200 000 g = 13.2 tonnes

Explanation

Industrial calculations follow the same steps as lab ones. The 2 : 1 ratio means ammonia runs out first even though there are more moles of it than of sulfuric acid.

Mark scheme

1 mark
moles of NH₃ = 3 400 000 ÷ 17 = 200 000 and moles of H₂SO₄ = 14 700 000 ÷ 98 = 150 000
allow 2.0 × 10⁵ and 1.5 × 10⁵; correct answer with no working scores full marks
1 mark
ammonia is limiting because 200 000 mol NH₃ needs only 100 000 mol H₂SO₄
allow sulfuric acid is in excess
1 mark
moles of (NH₄)₂SO₄ = 100 000 (mol)
allow ecf
1 mark
mass = 100 000 × 132 = 13 200 000 g = 13.2 tonnes
allow ecf; answer must be in tonnes

Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.

AI Tutor Hints first, answers after you try

Hi! Stuck? Ask me for a hint and I’ll help you work it out. Once you’ve checked your answer I can explain exactly where the marks are.