Combustion stoichiometry and molar volume at RTP Steps:

• Balance the combustion equation for complete reaction: C₁₂H₂₂O₁₁ + 6 O₂ → 12 CO₂ + 11 H₂O.
• The equation shows 1 mole of sugar (C₁₂H₂₂O₁₁) requires 6 moles of O₂.
• At RTP, the molar volume of any gas is 24 dm³ per mole.
• Calculate volume: 6 moles O₂ × 24 dm³/mol = 144 dm³.

Why C is correct:

• Stoichiometric ratio from balanced equation gives 6 moles O₂, multiplied by RTP molar volume (24 dm³/mol) yields 144 dm³.

Why the others are wrong:

• A: Equals 1 mole O₂ (24 dm³), ignores full reaction needs.
• B: Equals 1.5 moles O₂ (36 dm³), underestimates stoichiometry.
• D: Equals 9 moles O₂ (216 dm³), overestimates O₂ required.

Final answer: C