Combustion of glucose (C₆ H₁₂ O₆) produces CO₂ and water. The amount of oxygen (in g) required for the complete combustion of 900 g of glucose is :
[Molar mass of glucose in g mol⁻¹=180]

Solution

Chemistry ma’am · 4 steps

Balance first: each glucose burns with six O₂, so oxygen moles are six times glucose moles.

1. Balance the equation. Six C atoms give 6CO₂ and twelve H atoms give 6H₂O. That makes 18 O on the right. Glucose already brings 6 of them, so O₂ must supply the other 12, which is 6O₂.

   C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

2. Moles of glucose. n = 900/180 = 5 mol of glucose.
3. Mass of oxygen. The ratio is 1 : 6, so n_(O₂) = 6 × 5 = 30 mol. With M_(O₂) = 32 g mol⁻¹, the mass is 30 × 32 = 960 g.
4. So. Complete combustion of 900 g of glucose needs 960 g of oxygen, option (d).

The idea to keep. Grams to moles, then the mole ratio from the balanced equation, then moles back to grams. Oxygen gas is O₂, so its molar mass is 32, not 16.

Answer from the NTA final answer key. Solution by PrepWiser.