giúp mình câu 3

$
\begin{aligned}
& 8 \mathrm{Al}+30 \mathrm{HNO}_3 \rightarrow 8 \mathrm{Al}\left(\mathrm{NO}_3\right)_3+3 \mathrm{NH}_4 \mathrm{NO}_3+9 \mathrm{H}_2 \mathrm{O} \\
& 8 \mathrm{Al}+30 \mathrm{HNO}_3 \rightarrow 8 \mathrm{Al}\left(\mathrm{NO}_3\right)_3+3 \mathrm{~N}_2 \mathrm{O}+15 \mathrm{H}_2 \mathrm{O} \\
& 10 \mathrm{Al}+36 \mathrm{NNO}_3 \rightarrow 10 \mathrm{Al}\left(\mathrm{NO}_3\right)_3+3 \mathrm{~N}_2+18 \mathrm{H}_2 \mathrm{O} \\
& n_{\mathrm{Al}}=\frac{12,42}{27}=0,46(\mathrm{~mol}) \\
& n_Y=\frac{1,344}{22,4}=0,06(\mathrm{~mol}) \\
& \overline{M_Y}=18.2=36
\end{aligned}
$

Đặt $x, y$ là số mol $N_2 O, N_2$ trong $Y$. Ta có:
$
\begin{aligned}
& \left\{\begin{array}{l}
x+y=0,06 \\
44 x+28 y=0,06.36
\end{array}\right. \\
& \Rightarrow\left\{\begin{array}{l}
x=0,03 \\
y=0,03
\end{array}\right.
\end{aligned}
$

Giả sử dd sau phản ứng có $\mathrm{NH}_4 \mathrm{NO}_3$
Ta có: $n_{\mathrm{Al}}=\frac{8}{3} n_{\mathrm{NH}_4 \mathrm{NO}_3}+\frac{8}{3} n_{N_2 \mathrm{O}}+\frac{10}{3} n_{N_2}$
$\Rightarrow n_{N_4 H_4}=0,105(\mathrm{~mol})$ (giả sử đúng)
Ta có $n_{\mathrm{Al}\left(\mathrm{NO}_3\right)_3}=n_{\mathrm{Al}}=0,46(\mathrm{~mol})$
Vậy $m=0,46.213+0,105.80=106,38 g$