Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2024 — 8 April, Shift 1 — Question 59

Combustion of glucose $\mathrm{(C_6H_{12}O_6)}$ produces $\mathrm{CO_2}$ and water. The amount of oxygen (in g) required for the complete combustion of $900\ \mathrm{g}$ of glucose is:

[Given: Molar mass of glucose = $180\ \mathrm{g\ mol^{-1}}$ ]

Option A:
480
Option B:
960
Correct
Option C:
800
Option D:
32

Answer: B

Step-by-step solution

Balanced reaction: $\mathrm{C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O}$

Since, 1 mole glucose requires 6 mole $\mathrm{O_2}$ , thus

Moles of glucose $= \mathrm{\frac{900}{180} = 5\ mol}$

Required oxygen $= \mathrm{5 \times 6 = 30\ mol}$

Mass of oxygen $=\mathrm{30 \times 32 = 960\ g}$

Thus, the correct option is B.

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic: Chemical Equations, Stoichiometry and Limiting Reagent