How oxygen increased - AQA GCSE Chemistry Exam Questions

Practise 17 exam-style questions on How oxygen increased 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.

17 questions
1–6 marks each
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17 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

Write a word equation for photosynthesis.

Model answer

carbon dioxide + water → glucose + oxygen (light energy is needed)

Explanation

Photosynthesis uses carbon dioxide and water as reactants and produces glucose and oxygen. Light provides the energy, so it is written over the arrow, not as a reactant.

Mark scheme

1 mark
reactants: carbon dioxide + water
1 mark
products: glucose + oxygen
allow sugar for glucose; ignore light / chlorophyll written over the arrow

Question 2

2 marks

A student places some pondweed in water under a bright lamp. Bubbles of gas form on the leaves and are collected in a test tube.

Describe how the student could show that the gas contains a high proportion of oxygen.

Model answer

Insert a glowing splint into the test tube of gas. If the splint relights, the gas contains a high proportion of oxygen, produced by photosynthesis.

Explanation

This simple experiment shows that plants release oxygen in light, the same process that produced the oxygen in the atmosphere. The glowing splint test is the test for oxygen.

Mark scheme

1 mark
put a glowing splint into the test tube of gas
do not accept burning / lighted splint
1 mark
the splint relights

Question 3

3 marks

A plant takes in 4.4 g of carbon dioxide for photosynthesis.

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

Relative atomic masses (Ar): C = 12, O = 16
The volume of one mole of any gas at room temperature and pressure is 24 dm³.

Calculate the volume of oxygen produced at room temperature and pressure.

Model answer

Mr of CO₂ = 12 + (2 × 16) = 44
moles of CO₂ = 4.4 ÷ 44 = 0.10 mol
6 mol CO₂ gives 6 mol O₂, so moles of O₂ = 0.10 mol
volume of O₂ = 0.10 × 24 = 2.4 dm³

Explanation

The equation shows that photosynthesis releases one molecule of oxygen for every molecule of carbon dioxide taken in. Equal numbers of moles of gas occupy equal volumes, so the volume of oxygen released equals the volume of carbon dioxide absorbed.

Mark scheme

1 mark
moles of CO₂ = 4.4 ÷ 44 = 0.1(0) mol
1 mark
moles of O₂ = 0.1(0) mol
allow ecf; 6 : 6 = 1 : 1 ratio
1 mark
volume = 0.10 × 24 = 2.4 dm³
allow ecf; allow 2400 cm³; unit needed; correct answer with unit and no working scores 3 marks

Question 4

4 marks

A plant produces 90 g of glucose by photosynthesis.

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

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

Calculate the mass of oxygen produced.

Model answer

Mr of C₆H₁₂O₆ = (6 × 12) + (12 × 1) + (6 × 16) = 180
moles of glucose = 90 ÷ 180 = 0.50 mol
1 mol glucose is produced with 6 mol O₂, so moles of O₂ = 6 × 0.50 = 3.0 mol
mass of O₂ = 3.0 × 32 = 96 g

Explanation

Use the balanced equation to find the mole ratio between glucose and oxygen (1 : 6). Remember that oxygen gas is O₂, with Mr 32.

Mark scheme

1 mark
Mr of C₆H₁₂O₆ = 180
1 mark
moles of glucose = 90 ÷ 180 = 0.5(0) mol
allow ecf from an incorrect Mr
1 mark
moles of O₂ = 6 × 0.50 = 3(.0) mol
allow ecf
1 mark
mass of O₂ = 3.0 × 32 = 96 g
allow ecf; correct answer with no working scores 4 marks

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