12 litre of H 2 and 11.2 litre of Cl 2 are mixed and exploded. The volume of H 2 in the mixture is -

Question

12 litre of H 2 and 11.2 litre of Cl 2 are mixed and exploded. The volume of H 2 in the mixture is - (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

underset1 moleH2 + underset1 moleCl2 arrow underset2 mole2HCl 11.2 litre at NTP gas has =0.5 mole Cl 2 11.2=12 litre NTP =0.5 mole H 2 0.8 litre H 2= remaining 1 mole HCl =22.4 litre

Q. $12$ litre of $H_{2}$ and $11.2$ litre of $C l_{2}$ are mixed and exploded. The volume of $H_{2}$ in the mixture is -

$1 m o l e H_{2} + 1 m o l e C l_{2} \to 2 m o l e 2 H Cl$
$11.2$ litre at $NTP$ gas has $= 0.5$ mole $C l_{2}$
$11.2 = 12$ litre $NTP = 0.5$ mole $H_{2}$
$0.8$ litre $H_{2} =$ remaining
$1$ mole $H Cl = 22.4$ litre

Q. 12 litre of H2​ and 11.2 litre of Cl2​ are mixed and exploded. The volume of H2​ in the mixture is -

1moleH2​​+1moleCl2​​→2mole2HCl​ 11.2 litre at NTP gas has =0.5 mole Cl2​ 11.2=12 litre NTP=0.5 mole H2​ 0.8 litre H2​= remaining 1 mole HCl=22.4 litre

Q. $12$ litre of $H_{2}$ and $11.2$ litre of $C l_{2}$ are mixed and exploded. The volume of $H_{2}$ in the mixture is -

$1 m o l e H_{2} + 1 m o l e C l_{2} \to 2 m o l e 2 H Cl$
$11.2$ litre at $NTP$ gas has $= 0.5$ mole $C l_{2}$
$11.2 = 12$ litre $NTP = 0.5$ mole $H_{2}$
$0.8$ litre $H_{2} =$ remaining
$1$ mole $H Cl = 22.4$ litre