`M_X=0,5143.29=14,9147(g//mol)`

`->{n_{N_2\ bd}}/{n_{H_2\ bd}}={14,9147-2}/{28-14,9147}≈1`

`->\%V_{N_2\ bd}=\%V_{H_2\ bd}=50\%`

Cho `n_{N_2\ bd}=n_{H_2\ bd}=1(mol)`

`N_2+3H_2` $\xrightarrow{t^o,p,Fe}$ `2NH_3`

Đặt `n_{NH_3}=x(mol)`

Theo pt:

`n_{N_2\ pu}=0,5x(mol);n_{H_2\ pu}=1,5x(mol)`

`->n_{N_2\ du}=1-0,5x(mol);n_{H_2\ du}=1-1,5x(mol)`

`->M_Y={28(1-0,5x)+2(1-1,5x)+17x}/{1-0,5x+1-1,5x+x}=0,5517.29=16`

`->x=0,125(mol)`

`->n_{NH_3}=0,125(mol);n_{N_2\ du}=0,9375(mol);n_{H_2\ du}=0,8125(mol)`

`\%V_{NH_3}={0,125}/{0,125+0,9375+0,8125}.100\%=6,67\%`

`\%V_{N_2\ du}={0,9375}/{0,125+0,9375+0,8125}.100\%=50\%`

`\%V_{H_2\ du}=100-6,67-50=43,33\%`

`1>1/3->H` theo `H_2`

`H={0,8125}/{1}.100\%=81,25\%`