Combustion gas volume with water as liquid Steps:

• Reaction: 1 vol CH₄ + 2 vol O₂ → 1 vol CO₂ (+ 2H₂O(l), not gas).
• 10 cm³ CH₄ requires 20 cm³ O₂; 25 cm³ O₂ given, so 5 cm³ O₂ excess.
• CO₂ produced: 10 cm³ (1:1 with CH₄).
• Total gas volume: 10 cm³ CO₂ + 5 cm³ excess O₂ = 15 cm³. Why B is correct:
• Water forms liquid in standard combustion volume problems (Avogadro's law applies to gases only), so only CO₂ and excess O₂ count. Why the others are wrong:
• A: Ignores excess O₂; assumes complete consumption to just 10 cm³ CO₂.
• C: No basis; perhaps misadds volumes without limiting reactant.
• D: Assumes H₂O(g), adding 20 cm³ water vapor for 35 cm³ total. Final answer: B