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. $24$ $L$ of $HCl$
• B. $0.8$ $L$ $C{l}_2$ and $20.8$ $L$ $HCl$
• C. $0.8$ $L$ $H_2$ and $22.4$ $L$ $HCl$
• D. $22.4$ $L$ $HCl$ and $11.2$ $L$ $C{l}_2$

Answer 1

Answer: (C)

$0.8$ $L$ $H_2$ and $22.4$ $L$ $HCl$

Solution:- (C) $0.8LHCl$ and $22.4LHCl$

Given:-

Volume of $H_2=12L$

Volume of ${Cl}_2=11.2L$

$H_2+{Cl}_2⟶2HCl$

From the above reaction,

At STP, $22.4L$ of ${Cl}_2$ reacts with $22.4L$ mole of $H_2$ to give $44.8L$ mole of $HCl$.

Therefore,

$11.2L$ of ${Cl}_2$ reacts with $11.2L$ mole of $H_2$ to give $22.4L$ mole of $HCl$.

Volume of $H_2$ left after reacting with ${Cl}_2=12-11.2=0.8L$

Hence the composition by volume of mixture is $0.8LHCl$ and $22.4LHCl$.