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Liquids in Liquids

Fifty millili...

Question

Fifty millilitre of a mixture of CO and $C H_{4}$ was exploded with 85 ml of $O_{2}$ .The volume of $C O_{2}$ produced was 50 ml.Calculate the percentage composition of the gaseous mixture.

A

20%

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B

40%

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C

60%

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D

80%

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Solution

The correct option is B 80%
$2 C O + O_{2} \to 2 C O_{2}$
$C H_{4} + O_{2} \to C O_{2} + H_{2} O$

Assume $x$ mole of $C O$ participate in reaction. Then moles of $C H_{4}$ would be $50 - x$ mole.

From the reactions 2mole of $C O$ produce 2 mole of $C O_{2}$ . $x$ mol will produce $x$ mol of $C O_{2}$ . 2 mol of $C O$ requires 1 mol of $O_{2}$ . $x$ mol will require $x / 2$ mol $O_{2}$ .

1 mole $C H_{4}$ produces 1 mole $C O_{2}$ and requires 2 moles of $O_{2}$ . Therefore $x$ moles will produce $50 - x$ mole $C O_{2}$ and requires $2 (50 - x)$ mole $O_{2}$ .

Total mole of $O_{2}$ consumed = $x / 2 + 2 (50 - x)$
$200 - 3 x = 170$
$3 x = 30 \Rightarrow x = 10 m l$

Volume of $C O = 10 m l$
Volume of $C H_{4} = 50 - 10 = 40 m l$

percentage composition of $C H_{4} = \frac{40}{50} \times 100 = 80$ %

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