Q. When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, find the mass of aluminium oxide produced

Step 1: Write the balanced chemical equation
4Al + 3O₂ → 2Al₂O₃

Step 2: Calculate moles of reactants
Moles of Al = 81 / 27 = 3 moles
Moles of O₂ = 128 / 32 = 4 moles

Step 3: Identify the limiting reagent
For 3 moles of O₂, required Al = 4 moles
Available Al = 3 moles → Aluminium is the limiting reagent

Step 4: Calculate moles of Al₂O₃ formed
From equation: 4 moles Al → 2 moles Al₂O₃
So, 3 moles Al → (2/4) × 3 = 1.5 moles Al₂O₃

Step 5: Calculate mass of aluminium oxide
Molar mass of Al₂O₃ = (2×27) + (3×16) = 102 g mol⁻¹
Mass = 1.5 × 102 = 153 g