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Preparation of Chlorine

a g of KHC2...

Question

$a$ g of $K H C_{2} O_{4}$ is used to neutralize $100$ mL of $0.02 M K M n O_{4}$ in acid medium whereas $b$ g of $K H C_{2} O_{4}$ is used to neutralize $100$ mL of $0.02 M C a {(O H)}_{2}$ , then $x a = 5 b$ .
The value of $x$ is :

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Solution

The reaction is as follows:
$10 K H C_{2} O_{4} + 7 H_{2} S O_{4} + 4 K M n O_{4} \to 4 M n O + 20 C O_{2} + 12 H_{2} O + 7 K_{2} S O_{4}$
So, from the balanced reaction, it can be seen that $4 a = 5 b$ .
So, the value of $x$ is $4$ .

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