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Volume of Gases and Number of Moles

Potassium aci...

Question

Potassium acid oxalate $K_{2} C_{2} O_{4} . 3 H_{2} C_{2} O_{4} .4 H_{2} O$ can be oxidized by $M n O_{4}^{-}$ in acid medium. Calculate the volume in litresof $0.1 M K M n O_{4}$ reacting in acid solution with $5.08 g m$ of the acid oxalate.

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Solution

$K_{2} C_{2} O_{4} .3 H_{2} C_{2} O_{4} .4 H_{2} O + K M n O_{4} ⟶ M n^{+ 2} + 2 C O_{2}$ , The n factor for acid oxalate is $4$ and for $K M n O_{4}$ is $5$ ,
Now equate the gram equivalents-
Gram eq of $K M n O_{4}$ =Gram eq of acid oxalate
$N \times V = m o l e s \times V . f .$
$.5 \times V = \frac{5.08}{508.374} \times 4$
$V \approx .16$

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