KMnO4mol- Wt -158 oxidizes acid in acidic medium to CO2 and water as follows- 5C2O2-4-2MnO-16H- 10CO2-2Mn2-8H2OWhat is the equivalent weight of KMnO4-

The correct option is D 31.6 KMnO_4 molecular weight= 158 5C_2O_4^2 ++72MnO_4^ +16H^+⟶ 10CO_2++22Mn^2++8H_2O From above reaction we find t ...

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Equivalent Mass

KMnO4mol. Wt ...

Question

$K M n O_{4}$ (mol. Wt $= 158$ ) oxidizes acid in acidic medium to $C O_{2}$ and water as follows.
$5 C_{2} O_{4}^{2 -} + 2 M n O^{-} + 16 H^{+} \to 10 C O_{2} + 2 M n^{2 +} + 8 H_{2} O$
What is the equivalent weight of $K M n O_{4}$ ?

158

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31.6

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39.5

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79

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Solution

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K M n O_{4} (m . w . = 158)

C O_{2}

5 {C_{2} O_{4}}^{2 -} + 2 {M n O_{4}}^{-} + 16 H^{+} \to 10 C O_{2} + 2 M n^{2 +} + 8 H_{2} O

K M n O_{4}

K M n O_{4}

2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

K M n O_{4}

K M n O_{4}

2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

20 m L

K M n O_{4}

2 M n O_{4}^{-} + 16 H^{+} + 5 C_{2} O_{4}^{2 -} \to 2 M n^{2 +} + 8 H_{2} O + 10 C O_{2}

K M n O_{4} + 8 H^{+} + 5 e^{-} \to K^{+} + M n^{2 +} + 4 H_{2} O

K M n O_{4}

K M n O_{4} = 158

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