O Levels Chemistry (5070)•5070/11/M/J/22
Explanation
Stoichiometry of Al2O3 reduction to Al
Steps:
- Balanced equation: Al2O3 → 2Al + 3/2 O2 (1 mole Al2O3 yields 2 moles Al).
- Molar mass Al2O3 = 102 (2×27 + 3×16); 2Al = 54.
- Mass ratio Al2O3:Al = 102:54 = 1.89:1.
- For 27 tonnes Al: 27 × (102/54) = 51 tonnes Al2O3.
Why B is correct:
- 51 tonnes matches the stoichiometric mass ratio from molar masses in the reduction equation.
Why the others are wrong:
- A: 27 assumes equal masses, ignoring oxygen content in Al2O3.
- C: 54 tonnes Al2O3 produces 54 tonnes Al, double the required output.
- D: 102 tonnes Al2O3 produces 108 tonnes Al, quadruple the amount needed.
Final answer: B
Topic: Relative masses of atoms and molecules
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