Calculate moles of $K M n O_{4}$ required for complete oxidation of 2 moles of $K H C_{2} O_{4} . 2 H_{2} C_{2} O_{4}$ under acidic condition.

\frac{10}{5}

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\frac{6}{5}

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\frac{12}{5}

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\frac{8}{5}

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Solution

The correct option is C $\frac{12}{5}$
$K H C_{2} O_{4} . 2 H_{2} C_{2} O_{4}$
Each oxalate will lose 2 electrons. it has three oxalate in the formula
Hence, the total n-factor =6
n-factor for $K M n O_{4} = 5$
For a redox reaction,
number of equivalents of the oxidant = number of equivalents of the reductant

number of equivalents of the reductant $= moles \times n-factor = 2 \times 6 = 12$
number of equivalents of $K M n O_{4} = 12$
Hence moles of $K M n O_{4} = \frac{equivalents}{n-factor} = \frac{12}{5}$

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