Question

What mass of $MnO_2$ is reduced by 35 mL of 0.16 N oxalic acid in acidic solution? The skeleton equation is (Mn = 55) :- $MnO_2 + H^+ + H_2C_2O_4 \rightarrow CO_2 + H_2O + Mn^{2+}$

• A. 8.7 g
• B. 0.84 g
• C. 0.24 g
• D. 43.5 g

Answer 1

Answer: (C)

0.24 g

Answer 2

1. Balance the Redox Reaction: The balanced equation is: $MnO_2 + H_2C_2O_4 + 2H^+ \rightarrow Mn^{2+} + 2CO_2 + 2H_2O$. The stoichiometry between $MnO_2$ and $H_2C_2O_4$ is 1:1.

2. Calculate Equivalents of Oxalic Acid: Normality ($N$) of oxalic acid = 0.16 N. Volume ($V$) of oxalic acid = 35 mL = 0.035 L. Equivalents of $H_2C_2O_4 = N \times V = 0.16 \, N \times 0.035 \, L = 0.0056$ equivalents.

3. Equivalents of $MnO_2$: Since the stoichiometry is 1:1, Equivalents of $MnO_2$ = Equivalents of $H_2C_2O_4 = 0.0056$ equivalents.

4. Calculate Mass of $MnO_2$: Molar mass of $MnO_2 = 55 + (2 \times 16) = 87$ g/mol. The change in oxidation state for Mn in $MnO_2$ is from +4 to +2, so $n_f = 2$. Equivalent weight of $MnO_2 = \frac{\text{Molar mass}}{n_f} = \frac{87 \, g/mol}{2 \, eq/mol} = 43.5$ g/eq. Mass of $MnO_2 = \text{Equivalents} \times \text{Equivalent weight} = 0.0056 \, eq \times 43.5 \, g/eq = 0.2436$ g.

5. Select the Closest Option: The calculated mass is approximately 0.24 g.