Combustion stoichiometry of ethene with excess oxygen Steps:

• Balanced equation: C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l) at RTP.
• O₂ required for 100 cm³ C₂H₄ = 3 × 100 = 300 cm³.
• Excess O₂ remaining = 400 cm³ - 300 cm³ = 100 cm³.
• CO₂ produced = 2 × 100 cm³ = 200 cm³.
• Total final gas volume = excess O₂ + CO₂ = 100 cm³ + 200 cm³ = 300 cm³. Why C is correct:
• At RTP, H₂O condenses to liquid (negligible volume), leaving only CO₂ and excess O₂; volumes proportional to moles by Avogadro's law. Why the others are wrong:
• A: Subtracts all initial volumes without adding CO₂.
• B: Assumes incorrect 1:2.5 O₂ ratio or partial reaction.
• D: Assumes H₂O remains gaseous, preserving initial total volume ≈410 cm³ (rounded up). Final answer: C