Aqueous HCl reacts with MnO_ 2 (~s) to form MnCl_ 2 (aq), Cl_ 2 (~g) and H_ 2 O(l) . What is the weight (in g) of Cl_ 2 liberated when 8.7 g of MnO_ 2 (~s) is reacted with excess aqueous HCl solution? (Given Molar mass

Correct answer: D. 7.1 The balanced redox equation is: MnO_2 + 4HCl arrow MnCl_2 + Cl_2 + 2H_2O . Molar mass of MnO₂ = 55 + 32 = 87 g/mol. Moles of MnO₂ = (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 × 71 = 7.1 g.

JEE Main 202621 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Main 2026 · 21 January · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

Aqueous HCl reacts with MnO_ 2 (~s) to form MnCl_ 2 (aq), Cl_ 2...

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

)

Options

A21.3
B71
C14.2
D7.1

Correct answer

D. 7.1

Step-by-step 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

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