Equal no.of moles of $N a O H$ and acidic ${K M n O}_{4}$ are separately reacted with $N a H C_{2} O_{4}$ monosodium oxalate. Ratio idic of no. of moles of $N a H C_{2} O_{4}$ reacted with $N a O H$ and acidic ${K M n O}_{4}$ respectively is:

2 : 5

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5 : 2

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1 : 5

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None of these

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Solution

The correct option is A $2 : 5$
$N a O H + N a H C_{2} O_{4} \to N a_{2} C_{2} O_{4} + H_{2} O$
1 mole of $N a O H$ reacts with 1 mole of $N a H C_{2} O_{4}$ .
In acidic medium, $M n O_{4}^{-} + 5 e^{-} \to M n^{2 +}$
and $C_{2} O_{4}^{2 -} \to C O_{2} + 2 e^{-}$
On balancing we get, that 2 moles of $K M n O_{4}$ react with 5 moles of $N a H C_{2} O_{4}$ or 1 mole of $K M n O_{4}$ reacts with $\frac{5}{2}$ moles of $N a H C_{2} O_{4}$ .
Hence, required ratio $= 1 : \frac{5}{2} = 2 : 5$

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