AQA GCSE Chemistry Chemistry Paper 1 (Higher), November 2020: Question 6
9 marks · Standard Demand difficulty · Short Answer
Explain activation energy, determine the limiting reactant in the thermite reaction via mole calculations, complete an ionic displacement equation, and explain redox in terms of electron transfer.
Practise this questionQuestion
Question text
06 This question is about displacement reactions.
06.1 The displacement reaction between aluminium and iron oxide has a high
activation energy.
What is meant by ‘activation energy’?
[1 mark]
06.2 A mixture contains 1.00 kg of aluminium and 3.00 kg of iron oxide.
The equation for the reaction is:
2Al + Fe2O3 → 2Fe + Al2O3
Show that aluminium is the limiting reactant.
Relative atomic masses (Ar): O = 16 Al = 27 Fe = 56
[4 marks]
Magnesium displaces zinc from zinc sulfate solution.
06.3 Complete the ionic equation for the reaction.
You should include state symbols.
*16* [2 marks]
Mg(s) + Zn2+(aq) → … + …
06.4 Explain why the reaction between magnesium atoms and zinc ions is both oxidation
and reduction.
[2 marks]
Mark scheme
Show the mark scheme
Question 6
AO /
Question Answers Extra information Mark
Spec. Ref.
06.1 the (minimum) energy needed 1 AO1
for particles to react 4.5.1.2
or
the (minimum) energy needed allow the (minimum) energy
for a reaction to occur needed to start a reaction
06.2 (Mr of Fe2O3 =) 160 1 AO2
4.3.1.2
4.3.2.1
4.3.2.2
3000 4.3.2.4
(moles Fe2O3 = =)
18.75 (mol) allow correct use of incorrectly 1
calculated Mr
1000 allow 37.037037 (mol) correctly 1
(moles Al = =) 37.0 (mol)
rounded to at least 2 significant
figures
if both MP2 and MP3 are not
awarded allow 1 mark for
0.01875 mol Fe2O3 and 0.037
mol Al
(aluminium is limiting because) allow correct use of incorrect 1
37.0 mol is less than the (2 x number of moles from steps 2
18.75 =) 37.5 mol (aluminium and/or 3
needed)
or
iron oxide is in excess because
18.75 mol is more than the
37.0
( =) 18.5 mol (iron oxide
needed)
Question 6 continued
AO /
Spec. Ref.
06.2 alternative approaches: AO2
17 4.3.1.2
ctd
approach 1: 4.3.2.1
(finding required mass of 4.3.2.2
aluminium by moles method) 4.3.2.4
(Mr of Fe2O3 =) 160 (1)
3000
(moles Fe2O3 = =)
18.75 (mol) (1) allow correct use of incorrectly
calculated Mr
(moles Al needed
=18.75 × 2 = ) 37.5 (mol) allow correct use of incorrectly
and calculated moles of iron oxide
(mass Al needed = 37.5 × 27 =)
1012.5 (g) or 1.0125 kg (1) allow correct use of incorrectly
calculated moles of aluminium
needed
(so) 1.00 kg of aluminium is not dependent on calculated mass
enough (1) of aluminium needed being
greater than 1.00 (kg)
approach 2:
(finding required mass of
aluminium by proportion
method)
(Mr of Fe2O3 =) 160 (1)
(3.00 kg Fe2O3 needs)
3.00
× 2 × 27 (kg Al) (1) allow correct use of incorrectly
calculated Mr
(=) 1.0125 (kg) (1)
(so) 1.00 kg of aluminium is not dependent on calculated mass
enough (1) of aluminium needed being
greater than 1.00 (kg)
Question 6 continued
18 AO /
Spec. Ref.
06.2 alternative approaches: AO2
ctd 4.3.1.2
approach 3: 4.3.2.1
(finding required mass of iron 4.3.2.2
oxide by moles method) 4.3.2.4
Mr of Fe2O3 =) 160 (1)
1000
(moles Al = =) 37.0 (mol) allow 37.037037 (mol) correctly
rounded to at least 2 significant
(1)
figures
37.0
(moles Fe2O3 needed) = ) =
18.5 (mol) allow correct use of incorrectly
and calculated moles of aluminium
(mass Fe2O3 needed =
18.5 × 160 =) 2960 (g) or allow correct use of incorrectly
2.96 (kg) (1) calculated moles of iron oxide
needed
allow correct use of incorrectly
calculated Mr
(so) 3.00 kg of iron oxide is an dependent on calculated mass
excess (1) of iron oxide needed being less
than 3.00 (kg)
approach 4:
(finding required mass of iron
oxide by proportion method)
(Mr of Fe2O3 =) 160 (1)
1.00
(1.00 kg Al needs) ×160 allow correct use of incorrectly
2 x 27
(kg Fe2O3) (1) calculated Mr
(=) 2.96 (kg) (1)
(so) 3.00 kg of iron oxide is an dependent on calculated mass
excess (1) of iron oxide needed being less
than 3.00 (kg)
Question 6 continued19
AO /
Spec. Ref.
06.3 Mg(s) + Zn2+(aq) → allow multiples 2 AO2
Mg2+(aq) + Zn(s) 4.1.1.1
4.2.2.2
allow 1 mark for Mg2+ + Zn 4.4.1.4
with missing or incorrect state
symbols
06.4 magnesium (atoms) are 1 AO2
oxidised because they lose 4.4.1.4
electrons
(and) zinc (ions) are reduced 1
because they gain electrons
if no other marks awarded allow
1 mark for magnesium (atoms)
lose electrons and zinc (ions)
gain electrons
Total 9
How to answer it
Displacement Reactions, Limiting Reactants & Redox
This question assesses key skills across Quantitative Chemistry, Chemical Changes, and Energy Changes:
- Recalling the exact scientific definition of activation energy.
- Carrying out a multi-step limiting reactant calculation converting mass units (kg to g), finding moles ( n = m / Mᵣ ), and using stoichiometric molar ratios.
- Constructing balanced ionic equations with full state symbols for single displacement reactions.
- Explaining redox processes explicitly in terms of electron transfer ( OIL RIG ).
Defining Activation Energy
Topic: Energy Changes (Reaction Profiles)
✅ Mark Scheme Answer
The minimum energy needed for particles to react.
(Also accepted: the minimum energy needed for a reaction to occur / to start a reaction.)
🧠 Exam Technique
The word minimum is essential. Saying just "the energy needed to start a reaction" will lose the mark on many exam series because it does not state that it is a threshold barrier.
❌ Common Errors
- Omitting the word minimum (e.g. writing "the heat energy taken in to react").
- Confusing activation energy with overall energy change (ΔH).
💡 Key Knowledge
- Particles must collide with energy equal to or greater than the activation energy to have a successful collision.
- High activation energy means strong bonds must be broken before new bonds form, so the mixture often requires a spark or flame to initiate.
Limiting Reactant Calculation: Aluminium & Iron Oxide
Equation: 2Al + Fe₂O₃ → 2Fe + Al₂O₃
📐 Step-by-Step Calculation (Comparing Available Moles)
Mᵣ(Fe₂O₃) = (2 × 56) + (3 × 16) = 112 + 48 = 160 [Mark 1]
• Moles of Fe₂O₃ = 3000 g / 160 = 18.75 mol [Mark 2]
• Moles of Al = 1000 g / 27 = 37.0 mol (or 37.04 mol) [Mark 3]
From the equation: 2 moles of Al react with 1 mole of Fe₂O₃.
• To react all 18.75 mol of Fe₂O₃, required Al = 18.75 × 2 = 37.5 mol.
• We only have 37.0 mol of Al available (37.0 < 37.5).
Conclusion: Aluminium is the limiting reactant because there is not enough aluminium to react with all the iron oxide. [Mark 4]
❌ Common Calculation Traps
- Unit conversion: Forgetting that 1 kg = 1000 g. If using 1.00 and 3.00, units are kmol, but students often forget this and divide 1 / 27 directly without keeping units consistent.
- Ignoring the 2:1 ratio: Simply comparing 37.0 mol Al to 18.75 mol Fe₂O₃ and incorrectly claiming Fe₂O₃ is limiting because 18.75 is a smaller number. You must apply the reaction stoichiometry!
- Mᵣ errors: Calculating Mᵣ(Fe₂O₃) as (56 + 16) or using atomic numbers instead of mass numbers.
🧠 Alternative Method (Mass-Based)
You can also show Al is limiting by mass:
- 18.75 mol Fe₂O₃ needs 37.5 mol Al.
- Required mass of Al = 37.5 mol × 27 g/mol = 1012.5 g (1.0125 kg).
- Since 1.0125 kg is needed and only 1.00 kg is provided, aluminium runs out first and is the limiting reactant.
• Mark 1: Mᵣ of Fe₂O₃ = 160
• Mark 2: Moles of Fe₂O₃ = 3000 / 160 = 18.75 mol
• Mark 3: Moles of Al = 1000 / 27 = 37.0 mol
• Mark 4: Clear deduction comparing available moles to needed moles (37.0 < 37.5 mol).
Completing an Ionic Equation with State Symbols
Reaction: Magnesium + Zinc Sulfate
✅ Correct Completed Equation
Mg(s) + Zn²⁺(aq) → Mg²⁺(aq) + Zn(s)
Products can be written in either order: Mg²⁺(aq) + Zn(s) or Zn(s) + Mg²⁺(aq) .
💡 Key Knowledge: Why do spectator ions vanish?
In full aqueous solution:
Mg(s) + Zn²⁺(aq) + SO₄²⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq) + Zn(s)
Sulfate ions (SO₄²⁻) do not change state or charge; they are spectator ions and cancel out on both sides.
❌ Common Errors
- Missing out state symbols or writing incorrect ones (e.g., writing Mg²⁺(s) or Zn(aq)).
- Writing incorrect charges such as Mg⁺ or Zn⁺. Both magnesium and zinc form 2+ ions.
- Re-introducing the sulfate ion (SO₄²⁻) into an ionic equation.
🧠 Exam Technique
Read the question carefully: "You should include state symbols." In AQA mark schemes, this is worth an independent mark. Even if your formulae are correct, omitting state symbols caps your score at 1/2.
• 1 mark for correct species: Mg²⁺ + Zn
• 1 mark for correct state symbols: (aq) and (s)
Explaining Redox in Terms of Electrons
Linking Oxidation and Reduction
✅ Mark Scheme Answer
- Oxidation: Magnesium (atoms) are oxidised because they lose electrons. [Mark 1]
- Reduction: Zinc (ions) are reduced because they gain electrons. [Mark 2]
💡 The Golden Rule: OIL RIG
- Oxidation Is Loss of electrons:
Mg → Mg²⁺ + 2e⁻ - Reduction Is Gain of electrons:
Zn²⁺ + 2e⁻ → Zn
❌ Common Errors & Examiner Commentary
- Vague species identification: Saying "zinc is oxidised" instead of specifying zinc ions (Zn²⁺). In the reactant mixture, it is the Zn²⁺ ion gaining electrons, not zinc metal.
- Reverting to oxygen definitions: Explaining the reaction as "magnesium gains oxygen" when there is no oxygen in the ionic equation!
- Swapping the terms: Stating oxidation is gain of electrons. Always double-check with OIL RIG before writing.
🧠 Exam Technique
Structure your response into two distinct bullet points naming the exact species:
1. "Magnesium atoms lose 2 electrons to form Mg²⁺, so magnesium is oxidised."
2. "Zinc ions gain 2 electrons to form Zn atoms, so zinc ions are reduced."
• 1 mark for: magnesium (atoms) are oxidised because they lose electrons.
• 1 mark for: zinc (ions) are reduced because they gain electrons.
Topics
Chemistry · C3: Quantitative Chemistry · C4: Chemical Changes · C5: Energy Changes
Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 1 (Higher), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.