Em tham khảo nha:
\(\begin{array}{l}
2)\\
{n_{Mg}} = \dfrac{{2,4}}{{24}} = 0,1\,mol\\
{n_{{N_x}{O_y}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
Mg \to Mg + 2e\\
{N^{ + 5}} + (5 - \dfrac{{2y}}{x})e \to {N_x}{O_y}\\
BT\,e:2{n_{Mg}} = (5 - \dfrac{{2y}}{x}) \times {n_{{N_x}{O_y}}}\\
\Rightarrow 2 \times 0,1 = (5 - \dfrac{{2y}}{x}) \times 0,2\\
\Rightarrow 5 - \dfrac{{2y}}{x} = 1 \Leftrightarrow \dfrac{{2y}}{x} = 4 \Leftrightarrow 2y = 4x\\
\Rightarrow x:y = 1:2 \Rightarrow CTHH:N{O_2}\\
3)\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2\,mol\\
{n_{hh}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol \Rightarrow {n_{NO}} = {n_X} = \dfrac{{0,3}}{2} = 0,15\,mol\\
Fe \to F{e^{ + 3}} + 3e\\
{N^{ + 5}} + 3e \to NO\\
{N^{ + 5}} + (5 - n)e \to {N^{ + n}}\\
BT\,e:3{n_{Fe}} = 3{n_{NO}} + (5 - n) \times {n_X}\\
\Leftrightarrow 3 \times 0,2 = 3 \times 0,15 + (5 - n) \times 0,15\\
\Leftrightarrow n = 4 \Rightarrow X:N{O_2}\\
4)\\
{n_{Al}} = \dfrac{{6,48}}{{27}} = 0,24\,mol\\
{n_X} = \dfrac{{0,896}}{{22,4}} = 0,04\,mol\\
N{H_4}N{O_3} + NaOH \to NaN{O_3} + N{H_3} + {H_2}O\\
{n_{N{H_3}}} = \dfrac{{1,12}}{{22,4}} = 0,05\,mol \Rightarrow {n_{N{H_4}N{O_3}}} = {n_{N{H_3}}} = 0,05\,mol\\
Al \to A{l^{ + 3}} + 3e\\
{N^{ + 5}} + ne \to X\\
{N^{ + 5}} + 8e \to {N^{ - 3}}\\
BT\,e:3{n_{Al}} = {n_X} \times n + 8{n_{N{H_4}N{O_3}}}\\
\Leftrightarrow 3 \times 0,24 = n \times 0,04 + 8 \times 0,05\\
\Leftrightarrow n = 8 \Rightarrow X:{N_2}O
\end{array}\)
