Answer: C (144 dm³). The carbohydrate \((\text{CH}_2\text{O})_6\) has the molecular formula C₆H₁₂O₆. Burning it in excess oxygen: \(\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \to 6\text{CO}_2 + 6\text{H}_2\text{O}\).
From the balanced equation, 1 mole of the carbohydrate produces 6 moles of carbon dioxide gas. Only the carbon dioxide is collected as gas; the water forms as a liquid at room temperature and is not collected.
At room temperature and pressure the molar gas volume is 24 dm³, so the volume of CO₂ is \(6 \times 24 = 144\,\text{dm}^3\), which is option C. The most common error is to count the water vapour as a collected gas and add it in (giving 288 dm³, option D), but at r.t.p. the water is liquid, so only the 6 moles of carbon dioxide count.
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