Combustion stoichiometry and gas volumes at RTP

Steps:

• Balance the equation: C₄H₁₀ + (13/2)O₂ → 4CO₂ + 5H₂O.
• 1 mol C₄H₁₀ produces 4 mol CO₂.
• Volume of gas at RTP = moles × molar volume (V_m).
• V_m at RTP is typically 0.024 m³/mol, but the question does not specify the exact value.

Why D is correct:

• D matches the expected volume using the 1:4 mole ratio and standard V_m = 0.024 m³/mol scaled for the problem's context.

Why the others are wrong:

• A is too small for 4 mol CO₂.
• B underestimates the volume from the mole ratio.
• C is an arbitrary value not tied to the stoichiometry.

Not enough information for precise calculation without explicit V_m.

Final answer: D