In Haber process 30 liters of dihydrogen and 30 liters of dinitrogen were taken for reaction which

yielded only 50% of the expected product. What will be the composition of gaseous mixtureunder the aforesaid condition in the end.  

${\text{N}}_{\text{2}}{\text{\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3H}}_{\text{2}}\rightleftharpoons {\text{\hspace{0.17em}2NH}}_{\text{3}}$

$\text{30\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}30 \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0}$

$\text{30\hspace{0.17em}}-\text{\hspace{0.17em}x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}30\hspace{0.17em}}-\text{\hspace{0.17em}3x\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2x}$

20 L should come on complete reaction

$\therefore$50% to of it is 2x = 10; x = $\frac{10}{2}=5$

${\text{N}}_{\text{2}}\text{\hspace{0.17em}=\hspace{0.17em}30\hspace{0.17em}}-\text{\hspace{0.17em}5\hspace{0.17em}=\hspace{0.17em}25\hspace{0.17em}litre}$

${\text{H}}_{\text{2}}\text{\hspace{0.17em}=\hspace{0.17em}30\hspace{0.17em}}-\text{\hspace{0.17em}3\hspace{0.17em}×\hspace{0.17em}5\hspace{0.17em}=\hspace{0.17em}15\hspace{0.17em}litre}$

${\text{NH}}_{\text{3}}\text{\hspace{0.17em}=\hspace{0.17em}2\hspace{0.17em}×\hspace{0.17em}5\hspace{0.17em}=\hspace{0.17em}10\hspace{0.17em}litre}$