$1$ L of $C O_{2}$ is passed over red hot coke. The volume becomes $1.4$ L. The composition of the product(s) is:

0.6

C O

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0.8

C O_{2}

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0.6

C O_{2}

and

0.8

C O

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0.8

C O_{2}

and

0.6

L of

C O

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Solution

The correct option is C $0.6$ L $C O_{2}$ and $0.8$ L $C O$
Red hot coke reduces $C O_{2}$ to $C O$ . The balanced equation is $C O_{2} + C \to 2 C O$ .

$∴ 1$ mol of $C O_{2}$ will give $2$ mol of $C O$ .

$x$ L of $C O_{2}$ will give $2 x$ L of $C O$ .

In the question, $1$ L of $C O_{2}$ will react to give $1.4$ L of mixture.
Out of $1$ L, $x$ L of $C O_{2}$ reacts. The unreacted quantity is $1 - x$ .

Total volume of mixture $=$ unreacted $C O_{2} +$ volume of $C O$ formed.

Total volume of mixture $= 1 - x + 2 x = 1.4$ L.

Hence, $x = 0.4$ L.

Thus, volume of $C O$ formed = $2 \times 0.4 L$ $= 0.8 L$ .

Volume of $C O_{2}$ left = $(1 - 0.4) L = 0.6 L$ .

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1 L of CO2 is passed over red hot coke. The volume becomes 1.4 L. The composition of the product(s) is:

$1$ L of $C O_{2}$ is passed over red hot coke. The volume becomes $1.4$ L. The composition of the product(s) is: