a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)
PTHH:
4Al + 3O2 --to--> 2Al2O3
0,2-->0,15
2Mg + O2 --to--> 2MgO
1,2<--0,6
b) \(m_{Mg}=1,2.24=28,8\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{28,8}{28,8+5,4}.100\%=84,21\%\\\%m_{Al}=100\%-84,21\%=15,79\%\end{matrix}\right.\)