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Stoichiometric Calculations

A manna of ...

Question

A manna of $H C O O H$ and $H_{2} C_{2} O_{4}$ is heated with concentrated $H_{2} S O_{4}$ . The gas produced Is collected and on treating with KOH solution, the volume of the gas decreases by 1/6th. Calculate the molar ratio of the two acids in the original mixture.

2 : 1

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1 : 1

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4 : 1

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2 : 3

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Solution

The correct option is D $4 : 1$
$H C O O H 1 m o l C o n c . H_{2} S O_{4} \to C O 1 m o l + H_{2} O$

$H_{2} C_{2} O_{4} 1 m o l C o n c . H_{2} S O_{4} - ------- \to H e a t + C O 1 m o l + H_{2} O$

Let a moles of $H C O O H$ and b moles of $H_{2} C_{2} O_{4}$ be present in the original mixture.
Moles of $C O$ formed $= a + b$
Moles of $C O_{2}$ formed $= b$
Total moles of gases $= a + b + b = a + 2 b$
$C O_{2}$ is absorbed by $K O H$ and the volume reduces by $1 / 6 t h .$

Moles of $C O_{2} = \frac{a + 2 b}{6}$
or $b = \frac{a + 2 b}{6}$
or $a = 4 b$
or $\frac{a}{b} = 4$
or $a : b = 4 : 1$

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H_{2} C_{2} O_{4}

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H C O O H H_{2} S O_{4} - --- \to H_{2} O + C O

H_{2} C_{2} O_{4} H_{2} S O_{4} - --- \to H_{2} O + C O_{2} + C O

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