Q75 / 180 · AIIMS 2004 · Chemistry
For decolourization of 1 mole of $\mathrm{KMnO}_4$, the moles of $\mathrm{H}_2 \mathrm{O}_2$ required is
- A$1 / 2$
- B$3 / 2$
- Correct answerC$5 / 2$
- D$7 / 2$.
Solution
Acidified $\mathrm{KMnO}_4$ is decolourised by $\mathrm{H}_2 \mathrm{O}_2$ in following way.
$$
\begin{aligned}
& 2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+3 \mathrm{H}_2 \mathrm{O}+5 \mathrm{O} \\
& 5 \mathrm{H}_2 \mathrm{O}_2+5 \mathrm{O} \rightarrow 5 \mathrm{H}_2 \mathrm{O}+5 \mathrm{O}_2 \\
& 2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4 \\
& +8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{O}_2 \\
&
\end{aligned}
$$
Thus, for 2 moles of $\mathrm{KMnO}_4$ required $\mathrm{H}_2 \mathrm{O}_2=5$ For 1 mole of $\mathrm{KMnO}_4$ required $\mathrm{H}_2 \mathrm{O}_2=5 / 2$ mole.
Hence, the correct answer is option: 'C'.