When $C O_{2} (g)$ is passed over red-hot coke it partially gets reduced to $C O (g)$ . Upon passing $0.5$ litre of $C O_{2} (g)$ over red-hot coke, the total volume of the gases increased to $700 m L$ . The composition of the gaseous mixture at STP is :

C O_{2} = 200 m L; C O = 500 m L

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C O_{2} = 350 m L; C O = 350 m L

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C O_{2} = 0.0 m L; C O = 700 m L

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C O_{2} = 300 m L; C O = 400 m L

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Solution

The correct option is D $C O_{2} = 300 m L; C O = 400 m L$
$C O_{2} (g) + C (s) \to 2 C O (g)$

$0.5 L$

Given :

Total volume = $700 m l = 0.7 L$

$(0.5 - x) + (2 x) = 0.7$

$0.5 + x = 0.7$

$x = 0.2 L$ = $200 m l$

$C O_{2} (g) = 0.5 x = 0.5 - 0.2 = 300 m l$

$C O (g) = 2 x = 0.2 = 0.4 = 400 m l$

Therefore, the correct option is D.

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