Stoichiometry of alumina formation: If 1.5 moles of O2 react with aluminium to form Al2O3, what mass of aluminium (atomic weight = 27) is consumed?

Introduction / Context:
Simple stoichiometric calculations underpin material balances and reagent planning. Aluminium reacts with oxygen to form aluminium oxide (alumina), and the balanced equation provides the mole ratios needed to convert oxygen consumption into aluminium usage.

Given Data / Assumptions:

• Balanced reaction: 4 Al + 3 O2 → 2 Al2O3 (equivalently, 2 Al + 1.5 O2 → Al2O3).
• Moles of O2 consumed: 1.5 mol.
• Atomic weight of Al: 27 g/mol.

Concept / Approach:
Use the stoichiometric coefficients from the balanced equation to relate moles of oxygen to moles of aluminium, then multiply by the molar mass to find the mass of aluminium consumed.

Step-by-Step Solution:

Verification / Alternative check:
Using the 4:3:2 stoichiometry also yields Al:O2 = 4:3 in moles; 1.5 mol O2 × (4/3) = 2 mol Al, as above.

Why Other Options Are Wrong:

• 27 g would correspond to 1 mol Al, not enough for 1.5 mol O2.
• 5.4 g and 2.7 g are off by factors of 10.
• 81 g exceeds the stoichiometric requirement.

Common Pitfalls:
Failing to balance the reaction correctly or using atomic instead of molecular stoichiometry for oxygen (O2 vs. O).

Final Answer:
54 g