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Gay Lussac's Law, Avagadro's Law

20 ml of a mi...

Question

20 ml of a mixture of $C_{2} H_{2}$ and CO was exploded with 30ml of oxygen. The gases after the reaction had a volume of 34ml. On treatement with $K O H$ , 8ml of oxygen remained. Calculate the composition of the mixture.

A

C_{2} H_{2} = 6 m l, C O = 14 m l

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B

C_{2} H_{2} = 4 m l, C O = 28 m l

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C

C_{2} H_{2} = 8 m l, C O = 30 m l

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D

C_{2} H_{2} = 10 m l, C O = 19 m l

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Solution

The correct option is D $C_{2} H_{2} = 6 m l, C O = 14 m l$
Let the original composition of the mixture be $x$ ml $C_{2} H_{2}$ and $20 - x$ ml $C O$ .

The equations are as shown below:

$C_{2} H_{2} x + 2.5 O_{2} 2.5 x \to 2 C O_{2} 2 x + H_{2} O x$

$C O 20 - x + 0.5 O_{2} 0.5 (20 - x) \to C O_{2} 20 - x$

The volume of unreacted oxygen is 8 ml.

Hence, $30 m l - 2.5 x - 0.5 (20 - x) = 8 m l$ or $x = 6 m l$ .

Hence, the composition of the mixture is 6 ml $C_{2} H_{2}$ and 14 ml $C O$ .

Hence, the correct answer is option $A$ .

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