AQA A-Level Chemistry AS Paper 1, June 2023: Question 12
1 mark · Medium difficulty · Multiple Choice
Identify which reaction has the highest percentage atom economy for the production of hydrogen.
Practise this questionQuestion
Question text
12 Which reaction has the highest percentage atom economy for the production of
hydrogen?
[1 mark]
A LiH + H2O → LiOH + H2
B CO + H2O → CO2 + H2
C 2Al + 3H2O → Al2O3 + 3H2
D CH4 + H2O → CO + 3H2
Mark scheme
Show the mark scheme
12 D 1 (AO1) CH4 + H2O → CO + 3H2
How to answer it
Calculating Atom Economy for Hydrogen Production
This question assesses fundamental quantitative chemistry from physical and inorganic specifications, specifically:
- Understanding the definition and formula for percentage atom economy.
- Correctly accounting for stoichiometric coefficients in mass calculations.
- Using the periodic table to find relative formula masses ( Mr ).
- Quick comparison techniques to solve multiple-choice quantitative questions efficiently under timed conditions.
Question 12
Multiple Choice: Determining the reaction with highest atom economy
✅ Correct Answer
D: CH₄ + H₂O → CO + 3 H₂
💡 Key Knowledge
Atom Economy Formula:
% Atom Economy = (Mass of desired product / Total mass of all products) × 100
Note: By conservation of mass, the total mass of all products equals the total mass of all reactants.
📐 Step-by-Step Calculation Breakdown
In every option, the desired product is hydrogen (H₂). Calculate the percentage atom economy for each reaction:
- Option A: LiH + H₂O → LiOH + H₂
• Mass of desired product = 1 × 2.0 = 2.0
• Total mass of products = Mr(LiOH) + Mr(H₂) = (6.9 + 16.0 + 1.0) + 2.0 = 23.9 + 2.0 = 25.9
• % Atom Economy = (2.0 / 25.9) × 100 = 7.72% - Option B: CO + H₂O → CO₂ + H₂
• Mass of desired product = 1 × 2.0 = 2.0
• Total mass of products = Mr(CO₂) + Mr(H₂) = (12.0 + 32.0) + 2.0 = 44.0 + 2.0 = 46.0
• % Atom Economy = (2.0 / 46.0) × 100 = 4.35% - Option C: 2 Al + 3 H₂O → Al₂O₃ + 3 H₂
• Mass of desired product = 3 × 2.0 = 6.0
• Total mass of products = Mr(Al₂O₃) + 3 × Mr(H₂) = [(2 × 27.0) + (3 × 16.0)] + 6.0 = 102.0 + 6.0 = 108.0
• % Atom Economy = (6.0 / 108.0) × 100 = 5.56% - Option D: CH₄ + H₂O → CO + 3 H₂
• Mass of desired product = 3 × 2.0 = 6.0
• Total mass of products = Mr(CO) + 3 × Mr(H₂) = (12.0 + 16.0) + 6.0 = 28.0 + 6.0 = 34.0
• % Atom Economy = (6.0 / 34.0) × 100 = 17.65%
Conclusion: Option D has by far the highest atom economy at ~17.6%.
🧠 Exam Technique (MCQ Shortcut)
You don't need exact percentages to answer this in 60 seconds:
- Compare the ratio of desired mass to total waste mass:
• A: 2 / 24 ≈ 1/12
• B: 2 / 44 = 1/22
• C: 6 / 102 ≈ 1/17
• D: 6 / 28 ≈ 1/4.7 - Since 1/4.7 is much larger than 1/12, 1/17, or 1/22, Option D is immediately obvious without tedious long division!
❌ Common Errors & Traps
- Forgetting balancing numbers: In C and D, failing to multiply H₂ by 3 gives 2 / 34 (5.9%) instead of 6 / 34 (17.6%), leading students to incorrectly guess A.
- Confusing Atom Economy with Yield: Atom economy is theoretical and depends purely on the stoichiometry of the equation, regardless of how much product is experimentally obtained.
- Omitting desired product from denominator: The denominator must be the total mass of all products (or all reactants), not just the co-products/waste.
Topics
Physical Chemistry · 3.1.2 Amount of Substance
Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.