`N_2+3H_2` $\buildrel{{t^o,p,xt}}\over\rightleftharpoons$ `2NH_3`

`n_{N_2\ l t}={9,916}/{24,79}=0,4(mol);n_{H_2\ l t}={22,311}/{24,79}=0,9(mol)`

`0,4>{0,9}/3->H` theo `H_2`

`n_{NH_3\ l t}=2/3n_{H_2\ l t}=0,6(mol)`

`->H={3,4}/{0,6.17}.100\%=33,33\%`

$n_{N_2}$ `=` $\dfrac{9,916}{24,79}$ `=0,4` mol
$n_{H_2}$ `=` $\dfrac{22,311}{24,79}$ `=0,9` mol
`N_2 + 3H_2 -> 2NH_3`
Vì $\dfrac{0,4}{1}$ `>` $\dfrac{0,9}{3}$ `->` Tính theo `H_2`
Ta có `2/3`$n_{H_2}$ `=` $n_{NH_3}$ `=0,9 . 2/3 = 0,6` mol
$n_{NH_3}$ `=` $\dfrac{3,4}{17}$ `=0,2` mol
`-> H%` `=` $\dfrac{0,2}{0,6}$ `.100% = 33,33%`