\(\left\{{}\begin{matrix}n_{NO}+n_{N_2}=\dfrac{5,6}{22,4}\\30.n_{NO}+28.n_{N_2}=7,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{NO}=0,1mol\\n_{N_2}=0,15mol\end{matrix}\right.\)
\(Bte:x.n_M=3.n_{NO}+10.n_{N_2}\\ \Leftrightarrow x.\dfrac{16,2}{M}=1,8\)
x | 1 | 2 | 3 |
M | 9 | 18 | 27 |
=> M là Al