Em tham khảo nha:
\(\begin{array}{l}
{N_2} + 3{H_2} \to 2N{H_3}\\
{n_{{N_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
{n_{{H_2}}} = \dfrac{2}{2} = 1\,mol\\
{n_{{N_2}}} < \dfrac{{{n_{{H_2}}}}}{3} \Rightarrow {H_2} \text{ dư ,tính theo $N_2$ }\\
T{H_1}:H = 50\% \\
{n_{N{H_3}}} = 2{n_{{N_2}}} = 0,5\,mol\\
H = 50\% \Rightarrow {m_{N{H_3}}} = 0,5 \times 17 \times 50\% = 4,25g\\
T{H_2}:H = 30\% \\
{n_{N{H_3}}} = 2{n_{{N_2}}} = 0,5\,mol\\
H = 30\% \Rightarrow {m_{N{H_3}}} = 0,5 \times 17 \times 30\% = 2,55g
\end{array}\)