AQA GCSE Chemistry Chemistry Paper 1 (Higher), 2024: Question 7

15 marks · Standard Demand difficulty · Short Answer

This multi-part question assesses knowledge of metallic bonding, alloy structure, chemical equations, ionic ratios, percentage composition, and gas volume calculations involving iron.

Practise this question

Question

The image contains six sub-questions (07.1 to 07.6) about iron. It asks for descriptions of thermal conductivity in metals, the hardness of alloys, identification of a balanced chemical equation, ionic ratios in Fe3O4, percentage by mass calculation, and a multi-step calculation for the volume of CO2 gas produced from 40.0 kg of Fe2O3.
Question text

07 This question is about iron.

07.1 Iron is a metal.

Describe how iron conducts thermal energy.

[2 marks]

07.2 Pure iron is too soft for many uses.

Explain why mixing iron with other metals makes alloys which are harder than

pure iron.

[3 marks]

07.3 When iron reacts with chlorine, 0.12 mol of iron reacts with 0.18 mol of chlorine (Cl2).

Which is the correct equation for the reaction?

[1 mark]

Tick ( ) one box.

Fe + Cl2 → FeCl2

Fe + 3Cl2 → FeCl6

2Fe + Cl2 → 2FeCl

2Fe + 3Cl2 → 2FeCl3

The most common oxides of iron are Fe2O3 and Fe3O4

07.4 What is the ratio of the numbers of ions in Fe3O4?

[1 mark]

Tick ( ) one box.

2 Fe2+ : 1 Fe3+ : 4 O2–

1 Fe2+ : 2 Fe3+ : 4 O2–

3 Fe2+ : 4 O2–

3 Fe3+ : 4 O2–

07.5 Calculate the percentage (%) by mass of iron in Fe3O4

Relative atomic masses (Ar): O = 16 Fe = 56

[3 marks]

*22* Percentage by mass of iron = %

07.6 Fe2O3 reacts with carbon to produce carbon dioxide.

The equation for the reaction is:

2Fe2O3(s) + 3C(s) → 4 Fe(s) + 3CO2(g)

Calculate the volume of carbon dioxide gas at room temperature and pressure that is

produced from 40.0 kg of Fe2O3 using excess carbon.

Relative formula mass (Mr): Fe2O3 = 160

The volume of 1 mole of any gas at room temperature and pressure is 24 dm3.

[5 marks]

Volume of carbon dioxide = dm3

Mark scheme

Show the mark scheme The mark scheme provides the answers for each sub-question. It includes the scientific explanations for conductivity and alloys, the correct balanced equation, the ionic ratio, the percentage calculation steps (72.4%), and the multi-step calculation for gas volume (9000 dm3).

Question 7

AO /

Question Answers Extra information Mark

Spec. Ref.

07.1 (thermal) energy is transferred allow heat is transferred 1 AO1

4.2.1.5

by delocalised electrons 1 4.2.2.8

AO /

Spec. Ref.

07.2 allow (positive / metal) ions for AO1

atoms throughout 4.2.2.7

(the alloy / mixture has) different 1

sized atoms

(so the) layers are distorted 1

(so the) layers cannot easily allow (so the) atoms cannot 1

slide slide over each other

AO /

Spec. Ref.

07.3 2 Fe + 3Cl2 → 2FeCl3 1 AO2

4.3.1.1

4.3.2.3

AO /

Spec. Ref.

07.4 1 Fe2+ : 2 Fe3+ : 4 O2– 1 AO2

– – – 4.4.2.2

AO /

Spec. Ref.

07.5 (Mr Fe3O4 =) 232 1 AO2

4.3.1.2

3 × 56 168 1

(% Fe =) × 100 allow × 100

232 232

allow correct use of an

incorrectly determined Mr using

the values of Ar given in the

question

= 72.4 (%) allow 72.41379 correctly 1

rounded to at least 2 significant

figures

AO /

Spec. Ref.

a maximum of 4 marks can be

07.6 awarded for a method which AO2

determines and uses the volume 4.3.2.1

of iron oxide as a gas 4.3.2.2

(40.0 kg =) 40000 (g) 1

40 000 allow correct use of an 1

(moles Fe2O3 = =) 250 incorrectly converted or

unconverted mass

3 allow correct use of an 1

(moles CO2 = 250 × =) 375 incorrectly determined number

of moles of Fe2O3

(volume of CO2 =) 375 × 24 allow correct use of an 1

incorrectly determined number

of moles of CO2

= 9000 (dm3) 1

Total Question 7 15

How to answer it

Iron: Structure, Bonding, and Calculations

What this question tests

This question assesses your understanding of metallic bonding, the properties of alloys, and your ability to perform quantitative chemistry calculations including reacting moles, percentage mass, and gas volumes.

Part 07.1

Thermal Conductivity in Metals

💡 Key Knowledge

  • Metals consist of a lattice of positive ions.
  • They have a "sea" of delocalised electrons that are free to move throughout the structure.

✅ Correct Answer

  • (Thermal) energy is transferred... [1 mark]
  • ...by delocalised electrons. [1 mark]

🧠 Exam Technique

Whenever a question asks how a metal conducts heat or electricity, the answer almost always involves delocalised electrons moving through the structure. Don't just say "electrons"; use the word "delocalised" to secure the mark.

Part 07.2

Why Alloys are Harder than Pure Metals

✅ Correct Answer

  1. Alloys contain different sized atoms.
  2. This distorts the layers of atoms.
  3. The layers cannot slide over each other easily.

❌ Common Errors

Students often forget to mention the layers. Simply saying "the atoms can't move" is too vague. You must explain that the regular arrangement/layers are disrupted, preventing sliding.

🧠 Examiner Commentary

Top-level responses use a clear three-step logic: Different sizes → Distorted layers → No sliding. If you are asked to draw this, show a regular grid of circles with a few much larger or smaller circles mixed in to break the straight lines.

Parts 07.3 & 07.4

Reacting Ratios and Ion Ratios

07.3: Finding the Equation

📐 Calculation Step

Look at the moles given: 0.12 mol Fe and 0.18 mol Cl₂.

Divide both by the smallest number (0.12) to find the ratio:

Fe = 0.12 / 0.12 = 1

Cl₂ = 0.18 / 0.12 = 1.5

The ratio is 1 : 1.5, which is the same as 2 : 3.

Correct Box: 2 Fe + 3 Cl₂ → 2 FeCl₃

07.4: Ions in Fe₃O₄

💡 Key Knowledge

Fe₃O₄ (Magnetite) is a "mixed oxide" containing both Iron(II) and Iron(III) ions. To balance the 4 O²⁻ ions (total charge -8), you need:

  • One Fe²⁺ (charge +2)
  • Two Fe³⁺ (charge +6)
  • Total positive charge = +8

Correct Box: 1 Fe²⁺ : 2 Fe³⁺ : 4 O²⁻

Part 07.5

Percentage Mass Calculation

📐 Step-by-Step Guide

  1. Calculate total Mᵣ of Fe₃O₄:
    (3 × 56) + (4 × 16) = 168 + 64 = 232 [1 mark]
  2. Identify mass of Iron (Fe) only:
    3 × 56 = 168
  3. Calculate percentage:
    (168 / 232) × 100 = 72.4% [2 marks]

🧠 Exam Technique

Always show your working. Even if you get the final percentage wrong, you can get "Error Carried Forward" marks if your Mᵣ calculation was clear.

Part 07.6

Gas Volume Multi-Step Calculation

📐 Step-by-Step Calculation

  1. Convert mass to grams:
    40.0 kg = 40,000 g [1 mark]
  2. Calculate moles of Fe₂O₃:
    Moles = Mass / Mᵣ = 40,000 / 160 = 250 mol [1 mark]
  3. Use the molar ratio (2 : 3):
    From the equation: 2 moles of Fe₂O₃ produce 3 moles of CO₂.
    Moles of CO₂ = 250 × (3 / 2) = 375 mol [1 mark]
  4. Calculate volume of gas:
    Volume = Moles × 24 dm³
    Volume = 375 × 24 = 9000 dm³ [2 marks]

❌ Common Trap: Units

The biggest mistake here is forgetting to convert kg to g. Chemistry calculations almost always use grams. 40kg = 40,000g.

🧠 Examiner Insight

Students who correctly identified the 2:3 ratio usually scored highly. If you struggle with ratios, remember: (Amount you have / Number in equation) × Number you want.

Topics

Chemistry · C2: Bonding, Structure and the Properties of Matter · C3: Quantitative Chemistry

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 1 (Higher), 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.