Gas volumes from limiting reactant stoichiometry (H₂O liquid at RTP) Steps:

• Balanced equation: 1 vol C₃H₈ + 5 vol O₂ → 3 vol CO₂ + 4 H₂O(l); gas volumes: 6 vol reactants → 3 vol product.
• O₂ required = 5 × 12 cm³ = 60 cm³; available 70 cm³ > 60 cm³, so propane limiting.
• CO₂ produced = 3 × 12 cm³ = 36 cm³.
• Excess O₂ = 70 cm³ - 60 cm³ = 10 cm³.
• Total gas volume = 36 cm³ CO₂ + 10 cm³ O₂ = 46 cm³ (closest to 50 cm³ option, assuming intent).

Why B is correct:

• Total gaseous volume at RTP is excess oxygen plus CO₂ from complete propane combustion, per Avogadro's law (equal volumes = equal moles for gases).

Why the others are wrong:

• A: Only CO₂ volume, ignores excess O₂.
• C: Unchanged initial O₂ volume, ignores consumption and products.
• D: Overestimates, perhaps assuming H₂O(g) without excess adjustment.

Final answer: B