Question

Aqueous HCl reacts with \( MnO_2(s) \) to form \( MnCl_2(aq) \), \( Cl_2(g) \) and \( H_2O(l) \). 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 in g mol\(^{-1}\) : Mn = 55, Cl = 35.5, O = 16, H = 1)

• A. 21.3
• B. 71
• C. 14.2
• D. 7.1

Hint:

In many inorganic reactions, a 1:1 molar ratio exists between the oxidizing agent (like \( MnO_2 \)) and the simple diatomic product (like \( Cl_2 \)).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The reaction between manganese dioxide and hydrochloric acid is a redox reaction. We use the balanced equation to determine the molar ratio between the reactant and the product.

Step 2: Key Formula or Approach:
Balanced Equation: \( MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O \)
Moles \( = \frac{\text{Mass}}{\text{Molar Mass}} \)

Step 3: Detailed Explanation:
Molar mass of \( MnO_2 = 55 + 2 \times 16 = 87 \text{ g/mol} \).
Molar mass of \( Cl_2 = 2 \times 35.5 = 71 \text{ g/mol} \).
Number of moles of \( MnO_2 = \frac{8.7 \text{ g}}{87 \text{ g/mol}} = 0.1 \text{ mol} \).
From the balanced equation, 1 mole of \( MnO_2 \) yields 1 mole of \( Cl_2 \).
So, 0.1 mole of \( MnO_2 \) yields 0.1 mole of \( Cl_2 \).
Weight of \( Cl_2 = \text{moles} \times \text{molar mass} = 0.1 \text{ mol} \times 71 \text{ g/mol} = 7.1 \text{ g} \).

Step 4: Final Answer:
The weight of chlorine liberated is 7.1 g.