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JEE Chemistry 2026 Question with Solution

Aqueous HCl reacts with MnO2(s)MnO_2(s) to form MnCl2(aq)MnCl_2(aq), Cl2(g)Cl_2(g) and H2O(l)H_2O(l). What is the weight (in g) of Cl2Cl_2 liberated when 8.7 g8.7 \, \text{g} of MnO2(s)MnO_2(s) is reacted with excess aqueous HCl solution ? (Given Molar mass in g mol−1\text{g mol}^{-1} : Mn = 5555, Cl = 35.535.5, O = 1616, H = 11)

  • A

    21.321.3

  • B

    7171

  • C

    14.214.2

  • D

    7.17.1

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: MnO2MnO_2 reacts with excess aqueous HCl. Mass of MnO2=8.7 gMnO_2 = 8.7 \, \text{g}.

Find: Weight of Cl2Cl_2 liberated.

The balanced reaction is

MnO2+4HCl→MnCl2+Cl2+2H2OMnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O

Molar mass of MnO2MnO_2 is

55+2×16=87 g/mol55 + 2 \times 16 = 87 \, \text{g/mol}

Molar mass of Cl2Cl_2 is

2×35.5=71 g/mol2 \times 35.5 = 71 \, \text{g/mol}

Number of moles of MnO2MnO_2 is

8.787=0.1 mol\frac{8.7}{87} = 0.1 \, \text{mol}

From the balanced equation, 11 mole of MnO2MnO_2 yields 11 mole of Cl2Cl_2. Therefore, moles of Cl2Cl_2 formed are

0.1 mol0.1 \, \text{mol}

Mass of Cl2Cl_2 is

0.1×71=7.1 g0.1 \times 71 = 7.1 \, \text{g}

Therefore, the weight of chlorine liberated is 7.1 g7.1 \, \text{g}. The correct option is D.

Mole Ratio Approach

Given: MnO2MnO_2 is the limiting reagent and HCl is in excess.

Find: Mass of Cl2Cl_2 produced.

Use the mole relation from the balanced equation:

MnO2:Cl2=1:1MnO_2 : Cl_2 = 1 : 1

So the number of moles of Cl2Cl_2 formed is equal to the number of moles of MnO2MnO_2 consumed.

First calculate moles of MnO2MnO_2 from the given mass:

moles of MnO2=8.787=0.1\text{moles of } MnO_2 = \frac{8.7}{87} = 0.1

Now convert moles of Cl2Cl_2 to mass using its molar mass:

mass of Cl2=0.1×71=7.1 g\text{mass of } Cl_2 = 0.1 \times 71 = 7.1 \, \text{g}

Hence, the liberated chlorine weighs 7.1 g7.1 \, \text{g}.

Common mistakes

  • Using an incorrect stoichiometric ratio between MnO2MnO_2 and Cl2Cl_2. This is wrong because the balanced equation shows a 1:11:1 mole ratio. Always balance the reaction first and then read the mole relation from the coefficients.

  • Calculating the molar mass of MnO2MnO_2 incorrectly as only manganese mass or by adding oxygen once. This is wrong because MnO2MnO_2 contains two oxygen atoms, so its molar mass is 55+2×16=87 g/mol55 + 2 \times 16 = 87 \, \text{g/mol}.

  • Treating the answer as moles instead of mass. This is wrong because the question asks for weight in grams. After finding 0.1 mol0.1 \, \text{mol} of Cl2Cl_2, multiply by its molar mass 71 g/mol71 \, \text{g/mol}.

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