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

To solve the problem, we need to determine how much chlorine gas (\( Cl_2 \)) is liberated when a given mass of manganese dioxide (\( MnO_2 \)) reacts with excess hydrochloric acid (\( HCl \)). We'll use stoichiometry to find the amount of \( Cl_2 \) produced.

The balanced chemical equation for the reaction is: 

\(MnO_2(s) + 4HCl(aq) \rightarrow MnCl_2(aq) + Cl_2(g) + 2H_2O(l)\)

1. First, calculate the molar mass of \( MnO_2 \):
   • Molar mass of \( Mn = 55 \, \text{g/mol} \)
   • Molar mass of \( O_2 = 2 \times 16 \, \text{g/mol} = 32 \, \text{g/mol} \)
   • Total molar mass of \( MnO_2 = 55 + 32 = 87 \, \text{g/mol} \)
2. Convert 8.7 g of \( MnO_2 \) to moles:
   • \(n = \frac{8.7 \, \text{g}}{87 \, \text{g/mol}} = 0.1 \, \text{mol}\)
3. From the balanced equation, 1 mole of \( MnO_2 \) produces 1 mole of \( Cl_2 \). Therefore, 0.1 moles of \( MnO_2 \) will produce 0.1 moles of \( Cl_2 \).
4. Calculate the mass of \( Cl_2 \) produced:
   • Molar mass of \( Cl_2 = 2 \times 35.5 \, \text{g/mol} = 71 \, \text{g/mol} \)
   • Mass of \(\( Cl_2 = n \times \text{molar mass} = 0.1 \times 71 = 7.1  \text{g}\)

Thus, the weight of \( Cl_2 \) liberated is 7.1 g.