${KMnO}_{4}$ (mol.wt.= 158) oxidizes oxalic acid in acid medium to ${CO}_{2}$ and water as follows $5C_{2}O_{4}^{2 -} + 2{MnO}_{4}^{-} + 16H^{+} \rightarrow 10{CO}_{2} + 2{Mn}^{2 +} + 8H_{2}O$ What is the equivalent weight of ${KMnO}_{4}$?

Question

Dr. Abubacker · Accepted Answer

(b)
$5C_{2}O_{4}^{2 -} + \begin{matrix}
 + 7 \
2MnO_{4}^{-}
\end{matrix} + 16H^{+} ightarrow 10CO_{2} + 2Mn^{2 +} + 8H_{2}O$
Equivalent weight=$\frac{molecular\ weight}{change\ in\ oxidation\ number}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = \frac{158}{5} = 31.6$

Q. ${KMnO}_{4}$ (mol.wt.= 158) oxidizes oxalic acid in acid medium to ${CO}_{2}$ and water as follows $5C_{2}O_{4}^{2 -} + 2{MnO}_{4}^{-} + 16H^{+} \rightarrow 10{CO}_{2} + 2{Mn}^{2 +} + 8H_{2}O$ What is the equivalent weight of ${KMnO}_{4}$?

Explanation

Reviewed by Dr. Abubacker

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