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