Combustion stoichiometry and gas volume at room conditions Steps:

• Moles of propane = mass / molar mass = 9.81 g / 44 g mol⁻¹ = 0.223 mol.
• Balanced equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O, so 3 mol CO₂ produced per 1 mol C₃H₈.
• Moles of CO₂ = 3 × 0.223 mol = 0.669 mol.
• Volume of CO₂ = moles × molar volume at room conditions = 0.669 mol × 23.5 dm³ mol⁻¹ = 15.7 dm³. Why D is correct:
• Matches the 3:1 CO₂:C₃H₈ mole ratio from the balanced equation and uses 23.5 dm³ mol⁻¹ as the molar volume for gases at room conditions. Why the others are wrong:
• A: Uses 1:1 mole ratio and STP molar volume of 22.4 dm³ mol⁻¹ (0.223 × 22.4 ≈ 4.89 dm³).
• B: Uses 1:1 mole ratio and approximate room molar volume of 24 dm³ mol⁻¹ (0.219 × 24 ≈ 5.24 dm³).
• C: Uses correct 3:1 ratio but STP molar volume approximated as 22 dm³ mol⁻¹ (0.669 × 22 ≈ 14.7 dm³).

Final answer: D