Stoichiometry of propane combustion using gas volumes

Steps:

• Balanced equation: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
• Moles of propane: 12 dm³ / 24 dm³ mol⁻¹ = 0.5 mol (room conditions ≈ 24 dm³/mol)
• Moles of O₂ required: 0.5 mol × 5 = 2.5 mol
• Mass of O₂: 2.5 mol × 32 g mol⁻¹ = 80 g

Why B is correct:

• The 5:1 molar ratio in the balanced combustion equation, combined with Avogadro's law for gas volumes, yields exactly 80 g of O₂ for 0.5 mol propane.

Why the others are wrong:

• A: Assumes incorrect 4:1 O₂:propane ratio, yielding 64 g (rounded to 60 g).
• C: Calculates for 1 mol propane (120 g O₂), double the required amount.
• D: Uses 24 dm³/mol but for 1 mol propane with excess (160 g O₂).

Final answer: B