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Volume of Gases and Number of Moles

1 L of CO 2...

Question

$1 L$ of ${C O}_{2}$ is passed over hot coke. When the volume of the reaction mixture becomes $1.4 L$ , the composition of the reaction mixture is:

0.6 L

C O

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0.8 L

{C O}_{2}

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0.6 L

{C O}_{2}

and

0.8 L

C O

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None of these

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Solution

The correct option is C $0.6 L$ ${C O}_{2}$ and $0.8 L$ $C O$
$C O_{2} + C ⟶ 2 C O$

$1$ litre of $C O_{2} \equiv 1$ mole of coke $\equiv 2$ litres of $C O$ .

Let $x$ litres of $C O_{2}$ reacts to form $2 x$ litres of $C O$

$V_{(m i x t u r e)} = V_{(n e w C O_{2})} + V_{(C O)} = 1.4 L$

$V_{(n e w C O_{2})} = V_{(i n i t i a l C O_{2})} - V_{(r e a c t e d C O_{2})}$

$V_{C O} = 2 x$

$∴ V_{(n e w C O_{2})} = 1.4 - V_{C O}$

$(1 - x) = 1.4 - 2 x$

$2 x - x = 0.4$

$x = 0.4 L$

$∴$ New volume of $C O_{2} = 1 - x = 1 - 0.4$

$= 0.6 L$ of $C O_{2}$

& Volume of CO= $2 x = 2 \times 0.4 = 0.8 L$ of CO

Therefore, (C) is the answer.

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1 L of CO2 is passed over hot coke. When the volume of the reaction mixture becomes 1.4 L, the composition of the reaction mixture is:

$1 L$ of ${C O}_{2}$ is passed over hot coke. When the volume of the reaction mixture becomes $1.4 L$ , the composition of the reaction mixture is: