One litre mixture of $C O$ and $C O_{2}$ is passed through red-hot charcoal in a tube. The new volume becomes 1.4 L. Fine out the composition of the mixture by volume.

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1L of a mixture of CO and CO2 becomes 1.6L when passed through a red hot charcoal, CO2 in the mixture

The correct option is D 60% Let 1L of CO and CO2 mixture contain x litres of CO2 C+CO2⟶2CO 1 mole 2 mole x litre 2x litre Therefore, total volume=(1+x)litres. 1+x=1.6, or, x=0.6 litres. Hence, 60% of CO2

1L of a mixture of CO and $C O_{2}$ becomes 1.6L when passed through a red hot charcoal, $C O_{2}$ in the mixture

The correct option is D
$60 %$

Let 1L of CO and $C O_{2}$ mixture contain x litres of $C O_{2}$
$C + C O_{2} ⟶ 2 C O$
1 mole 2 mole
x litre 2x litre
Therefore, total volume=(1+x)litres.
1+x=1.6, or, x=0.6 litres.
Hence, $60 % o f C O_{2}$