Combustion volume change using gas equation at constant T and P

Steps:

• Balanced equation: C₂H₆(g) + 3.5 O₂(g) → 2 CO₂(g) + 3 H₂O(g) (water vapor at 200°C).
• O₂ required: 3.5 × 10 cm³ = 35 cm³; excess O₂: 45 - 35 = 10 cm³.
• Gaseous products: (2 + 3) × 10 cm³ = 50 cm³.
• Total gas volume: excess O₂ + products = 10 + 50 = 60 cm³.

Why D is correct:

• D equals the total gaseous volume from Avogadro's law, where volumes are proportional to moles at constant T and P.

Why the others are wrong:

• A: Assumes H₂O(l), yielding only 20 cm³ CO₂ + 10 cm³ excess O₂ = 30 cm³.
• B: Counts products (50 cm³) but omits excess O₂.
• C: Arises from arithmetic error, like 45 - 35 + 45 (wrong O₂ use).

Final answer: D