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Stoichiometric Calculations

Mass of KHC...

Question

Mass of $K H C_{2} O_{4}$ (potassium acid oxalate) required to reduce $100$ mL of $0.02 M K M n O_{4}$ in acidic medium (to $M n^{2 +})$ is $x$ g and to neutralise $100$ mL of $0.05 M C a (O H)_{2}$ is $y$ g, then :

x = y

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2 x = y

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x = 2 y

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none of the above

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Solution

The correct option is D $2 x = y$
The reaction is as follows:
$5 {H C_{2} O_{4}}^{-} + 2 {M n O_{4}}^{-} \to 2 M n^{2 +} + 10 C O_{2}$
The number of moles of ${M n O_{4}}^{-}$ $= 0.1 \times 0.02 = 0.002$ mol
This is equal to $0.005$ mol of ${H C_{2} O_{4}}^{-}$ ion. This corresponds to $x$ g.
The number of moles of $C a (O H)_{2}$ is $0.1 \times 0.05 = 0.005$ mol
They corresponds to $0.01$ mol of hydrogen ions and $0.01$ mol of ${H C_{2} O_{4}}^{-}$ .
This is equal to $y$ g.
Hence, $\frac{x}{y} = \frac{0.005}{0.01} = \frac{1}{2}$ or $y = 2 x$
Hence, the relation is $2 x = y$ .

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