What mass of aluminum oxide and hydrogen iodide would be needed to produce 102g of aluminum iodide crystals?

Molar mass of Al₂O₃ = (27.0×2 + 16.0×3) g/mol = 102.0 g/mol
Molar mass of HI = (1.0 + 126.9) g/mol = 127.9 g/mol
Molar mass of AlI₃ = (27.0 + 126.9×3) g/mol = 407.7 g/mol

The balanced equation should be: Al₂O₃ + 6HI → 2AlI₃ + 3H₂O
OR: Mole ratio Al₂O₃ : HI : AlI₃ = 1 : 6 : 2

No. of moles of AlI₃ produced = (102 g) / (407.7 g/mol) = 0.250 mol

No. of moles of Al₂O₃ needed = (0.250 mol) × (1/2) = 0.125 mol
Mass of Al₂O₃ needed = (0.125 mol) × (102.0 g/mol) = 12.8 g

No. of moles of HI needed = (0.250 mol) × (6/2) = 0.750 mol
Mass of HI needed = (0.750 mol) × (127.9 g/mol) = 95.9 g