a,
$n_{N_2}=\dfrac{4,48}{22,4}=0,2(mol)$
$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$
$n_{\text{spu}}=\dfrac{7,84}{22,4}=0,35(mol)$
$N_2+3H_2\rightleftharpoons 2NH_3$
$\dfrac{0,3}{3}<\dfrac{0,2}{1}$
Theo lí thuyết $H_2$ hết, $N_2$
Thực tế $H_2$ phản ứng $x$ mol
Hỗn hợp sau phản ứng gồm:
$H_2: 0,3-x (mol)$
$N_2: 0,2-\dfrac{x}{3}(mol)$
$NH_3: \dfrac{2x}{3}(mol)$
$\to 0,3-x+0,2-\dfrac{x}{3}+\dfrac{2}{3}x=0,35$
$\to x=0,225(mol)$
$\to H=\dfrac{0,225.100}{0,3}=75\%$
b,
$V_{NH_3}=22,4.\dfrac{2x}{3}=3,36l$