Q15 / 75 · JEE Main 2026 (21 January Shift 2) · Chemistry
Aqueous HCl reacts with $\mathrm{MnO}_{2}(\mathrm{~s})$ to form $\mathrm{MnCl}_{2}(\mathrm{aq}), \mathrm{Cl}_{2}(\mathrm{~g})$ and $\mathrm{H}_{2} \mathrm{O}(l)$. What is the weight (in g) of $\mathrm{Cl}_{2}$ liberated when 8.7 g of $\mathrm{MnO}_{2}(\mathrm{~s})$ is reacted with excess aqueous HCl solution?
(Given Molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{Mn}=55, \mathrm{Cl}=35.5, \mathrm{O}=16, \mathrm{H}=1$)
(Given Molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{Mn}=55, \mathrm{Cl}=35.5, \mathrm{O}=16, \mathrm{H}=1$)
- A21.3
- B71
- C14.2
- Correct answerD7.1
Solution
The balanced redox equation is: $\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}$.
Molar mass of MnO₂ = 55 + 32 = 87 g/mol.
Moles of MnO₂ = $\frac{8.7}{87} = 0.1$ mol.
From stoichiometry, 1 mol MnO₂ produces 1 mol Cl₂, so 0.1 mol produces 0.1 mol Cl₂.
Mass of Cl₂ = $0.1 \times 71 = 7.1$ g.
Molar mass of MnO₂ = 55 + 32 = 87 g/mol.
Moles of MnO₂ = $\frac{8.7}{87} = 0.1$ mol.
From stoichiometry, 1 mol MnO₂ produces 1 mol Cl₂, so 0.1 mol produces 0.1 mol Cl₂.
Mass of Cl₂ = $0.1 \times 71 = 7.1$ g.
Hence, the correct answer is option: 'D'.