Question

$40$% mixutre of $0.2$ mole of $N_2$ and $0.6$ mole of $H_2$ react to give ${NH}_3$ according to the equation
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$
at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is _______.

• A. $4:5$
• B. $5:4$
• C. $7:10$
• D. $8:5$

Answer 1

The correct option is C $7:10$

$N_2+3H_2\to 2N{H}_3$
$0.2$ $0.6$
$0.2-X$ $0.6-3X$ $2X$
According to question$(0.2-x)+(0.6-3x)=0.4x0.8$
from here x coming out to be$0.12$
final volume will be$(0.08+0.24+0.24)=0.56moles$
initial volume is$(0.2+0.6)=0.8$

Ratio=$0.56/0.8=7/10=7:10$