Amounts of substances in equations (HT only) - AQA GCSE Chemistry Exam Questions

Practise 21 exam-style questions on Amounts of substances in equations (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.

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1–5 marks each
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20 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

Calcium reacts with water:

Ca + 2H₂O → Ca(OH)₂ + H₂

Describe what this equation shows in terms of moles.

Model answer

One mole of calcium reacts with two moles of water to produce one mole of calcium hydroxide and one mole of hydrogen gas.

Explanation

The numbers in front of the formulae in a balanced equation give the ratio of moles of each substance. Where there is no number, it means one mole.

Mark scheme

1 mark
1 mole of calcium reacts with 2 moles of water
1 mark
to produce 1 mole of calcium hydroxide and 1 mole of hydrogen
allow 'one mole of each product'

Question 2

3 marks

Magnesium burns in oxygen:

2Mg + O₂ → 2MgO

Calculate the mass of magnesium oxide formed when 6.0 g of magnesium burns completely.

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

Model answer

moles of Mg = 6.0 ÷ 24 = 0.25 mol
The equation shows 2 mol Mg gives 2 mol MgO, so moles of MgO = 0.25 mol
Mr of MgO = 24 + 16 = 40
mass of MgO = 0.25 × 40 = 10 g

Explanation

The three steps for a reacting mass calculation are: mass to moles of the known substance, use the equation's ratio to find moles of the unknown, then moles to mass. Here the ratio Mg : MgO is 2 : 2, which is 1 : 1.

Mark scheme

1 mark
moles of Mg = 6.0 ÷ 24 = 0.25 (mol)
1 mark
moles of MgO = 0.25 (mol) (1 : 1 ratio)
allow ecf
1 mark
mass of MgO = 0.25 × 40 = 10 g
allow ecf; correct answer with no working scores 3 marks; allow ratio method: 48 g Mg gives 80 g MgO

Question 3

3 marks

Ethanol burns in oxygen:

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

Calculate the mass of oxygen needed to burn 23 g of ethanol completely.

Relative atomic masses (Ar): H = 1, C = 12, O = 16

Model answer

Mr of C₂H₅OH = (2 × 12) + (6 × 1) + 16 = 46
moles of ethanol = 23 ÷ 46 = 0.50 mol
Ratio C₂H₅OH : O₂ is 1 : 3, so moles of O₂ = 1.5 mol
mass of O₂ = 1.5 × 32 = 48 g

Explanation

Count all six hydrogen atoms in C₂H₅OH when working out its Mr. Then use the 1 : 3 ratio from the equation, and remember that oxygen gas is O₂ with Mr 32.

Mark scheme

1 mark
Mr of C₂H₅OH = 46 and moles = 23 ÷ 46 = 0.50 (mol)
1 mark
moles of O₂ = 0.50 × 3 = 1.5 (mol)
allow ecf
1 mark
mass of O₂ = 1.5 × 32 = 48 g
allow ecf; correct answer with no working scores 3 marks; 24 g (using Mr of O = 16) scores 2 marks

Question 4

4 marks

A student heated different masses of zinc carbonate until each sample had completely decomposed, and weighed the zinc oxide left. The table shows the results.

ZnCO₃ → ZnO + CO₂

Use the equation to calculate the expected mass of zinc oxide from 3.75 g of zinc carbonate. Identify the anomalous result and suggest a reason for it.

Relative atomic masses (Ar): C = 12, O = 16, Zn = 65

Table with columns "Mass of zinc carbonate in g", "Mass of zinc oxide in g". Mass of zinc carbonate in g: 1.25; Mass of zinc oxide in g: 0.81 | Mass of zinc carbonate in g: 2.50; Mass of zinc oxide in g: 1.62 | Mass of zinc carbonate in g: 3.75; Mass of zinc oxide in g: 2.80 | Mass of zinc carbonate in g: 5.00; Mass of zinc oxide in g: 3.24.

Model answer

Mr of ZnCO₃ = 65 + 12 + (3 × 16) = 125; Mr of ZnO = 65 + 16 = 81
moles of ZnCO₃ = 3.75 ÷ 125 = 0.030 mol, so moles of ZnO = 0.030 mol
expected mass of ZnO = 0.030 × 81 = 2.43 g
The result for 3.75 g (2.80 g) is anomalous because it is higher than expected. The sample was probably not heated for long enough, so some zinc carbonate had not decomposed.

Explanation

The other results all fit the ratio 125 g of zinc carbonate to 81 g of zinc oxide. An anomalously high residue suggests unreacted starting material, because carbon dioxide still has to be lost.

Mark scheme

1 mark
moles of ZnCO₃ = 3.75 ÷ 125 = 0.030 (mol)
correct answer with no working scores full marks
1 mark
expected mass of ZnO = 0.030 × 81 = 2.43 g
allow ecf
1 mark
the result for 3.75 g (2.80 g) is anomalous
1 mark
not all the zinc carbonate decomposed / not heated for long enough
allow not heated to constant mass

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