Stoichiometric mass ratio from Al2O3 decomposition

Steps:

• The reaction is 2Al2O3 → 4Al + 3O2, so 1 mole Al2O3 produces 2 moles Al.
• Molar mass Al2O3 = 102 g/mol; molar mass of 2Al = 54 g/mol.
• Mass ratio Al2O3:Al = 102:54, or 102/54 = 1.89 (simplifies to 51/27).
• For 27 tonnes Al, Al2O3 required = 27 × (102/54) = 27 × (51/27) = 51 tonnes.

Why B is correct:

• The 51:27 mass ratio follows from the chemical formula of Al2O3, where oxygen mass (48 u) requires extra Al2O3 to yield pure Al.

Why the others are wrong:

• A: Assumes equal masses, ignoring oxygen in Al2O3.
• C: Matches atomic mass of Al (27×2), but overlooks full Al2O3 molar mass.
• D: Doubles the correct amount, possibly confusing moles of Al2O3 with Al atoms.

Final answer: B