- A$1,2,2$
- B$2,2,1$
- ✓$1,1,2$
- D$2,1,2$
at $t = 0$ $56\,gm$ $8\,gm$ $0\,gm$
$=$ $2\,mole$ $4\,mole$ $0\,mole$
at equilibrium $2 - 1$ $4 - 3$ $34\,gm$
$=$ $1\,mole$ $=$ $1\,mole$ $=$ $2\,mole$
According to eq. $(i)$ $2$ $mole$ of ammonia are present $\&$ to produce $2$ $mole$ of $N{H_3}$, we need $1$ mole of ${N_2}$ and $3$ $mole$ of ${H_2}$ hence $2 - 1 = 1$ $mole$ of ${N_2}$ and $4 - 3 = 1$ mole of ${H_2}$ are present at equilibrium in vessel.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

$56\ g$ of nitrogen and $8\ g$ of hydrogen gas are heated in a closed vessel. At equilibrium, $34\ g$ of ammonia are present. The equilibrium number of moles of nitrogen, hydrogen and ammonia are respectively