Question

12 litre of $H_2$ and 11.2 litre of $C{l}_2$ are mixed and exploded. The composition by volume of mixture is:

• A. 24 litre of $HCl$
• B. 0.8 litre $C{l}_2$ and 20.8 lit $HCl$
• C. 0.8 litre $H_2$ and 22.4 litre $HCl$
• D. 22.4 litre $HCl$

Answer 1

Answer: (C)

0.8 litre $H_2$ and 22.4 litre $HCl$

The balanced equation for this reaction is :

$H_2+{Cl}_2\to 2HCl$

1 mole of $H_2$ reacts with 1 mole of ${Cl}_2$ to give 2 moles of $HCl$.

Since 22.4 litres of a gas occupies 1 mole.

Hence, moles of 12 L of $H_2$ will be $=\frac{12}{22.4}=0.54$ moles

Similarly moles of 11.2 L ${Cl}_2$ gas $=\frac{11.2}{22.4}=0.5$ moles

$C{l}_2$ is the limiting reagent it is present in a limited amount.

Therefore, 0.5 moles of the $C{l}_2$ will give 1 mole of $HCl$.

Since $H_2$ was in excess, hence some part of it is unreacted.

So moles of unreacted $H_2$ is $\frac{12}{22.4}-\frac{11.2}{22.4}=0.035$ moles

Volume of unreacted $H_2$ is $=0.035\times 22.4=0.8$ litres

Therefore, the composition of the volume of mixtures is 22.4 litres of $HCl$ and 0.8 litres of $H_2$.