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Volume of Gases and Number of Moles

20 ml of mixt...

Question

20 ml of mixture of $C O$ and $C_{2} H_{4}$ was exploded with 50 mL of $O_{2}$ . The volume after explosion was 45 mL of shaking with $N a O H$ solution only 15 mL $O_{2}$ was left behind , what were the volume of $C O$ and $C_{2} H_{4}$ in mixture?

A

C O = 5 m L, C_{2} H_{2} = 15 m L

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B

C O = 10 m L, C_{2} H_{2} = 10 m L

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C

C O = 8 m L, C_{2} H_{2} = 12 m L

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D

C O = 15 m L, C_{2} H_{2} = 5 m L

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Solution

The correct option is B $C O = 10 m L, C_{2} H_{2} = 10 m L$
Solution:-
Let the mixture contain $x m L$ of $C_{2} H_{4}$ and $y m L$ of $C O$ .

$C_{2} H_{4} x + 3 O_{2} ⟶ 2 C O_{2} 2 x + 2 H_{2} O$

$C O y + \frac{1}{2} O_{2} ⟶ C O_{2} y$

$N a O H$ absorbs the $C O_{2}$ produced in the reaction.

Total amount of $C O_{2}$ produced $= 2 x + y = (45 - 15)$
$\Rightarrow 2 x + y = 30 . . . . . (1)$

Given that:-

$x + y = 20 . . . . . (2)$
From ${e q}^{n} (1) & (2)$ , we have

$20 + y = 30$
$\Rightarrow y = 10 m L$

Substituting the value of $y$ in ${e q}^{n} (2)$ , we have
$x + 10 = 20$
$\Rightarrow x = 10 m L$
Hence the volume of $C O$ and $C_{2} H_{4}$ in the mixture is $10 m L$ and $10 m L$ respectively.

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