Phải là \(N_2\) nhé
\(\begin{array}{l}
n_{N_2}=\frac{11,2}{28}=0,4(mol)\\
n_{H_2}=\frac{3}{2}=1,5(mol)\\
N_2+3H_2\buildrel{{t^o,xt,p}}\over\rightleftharpoons 2NH_3\\
Vi\,n_{N_2}<\frac{n_{H_2}}{3}\to H\,tinh\,theo\,N_2\\
Theo\,PT:\,n_{NH_3(lt)}=2n_{N_2}=0,8(mol)\\
\to H=\frac{3,4}{0,8.17}.100\%=25\%
\end{array}\)