Giải thích các bước giải:
a) `PT:`
$4Al+3O_2\xrightarrow{t^0}2Al_2O_3$
$2Mg+O_2\xrightarrow{t^0}2MgO$
b) Gọi `n_(Al)=a(mol);n_(Mg)=b_(mol)`
`->27a+24b=6,6` $(1)$
Theo PT có:
`n_(Al_2O_3)=1/2 n_(Al)=1/2a(mol)`
`n_(MgO)=n_(Mg)=b(mol)`
`->102. 1/2a+40b=51a+40b=12,2` $(2)$
Từ `(1);(2)->`$\left\{\begin{matrix}27a+24b=6,6\\51a+40b=12,2\end{matrix}\right.$`->`$\left\{\begin{matrix}a=0,2(mol)\\b=0,05(mol)\end{matrix}\right.$
`->`$\left\{\begin{matrix}m_{Al}=27×0,2=5,4(g)\\m_{Mg}=24×0,05=1,2(g)\end{matrix}\right.$
`->`$\left\{\begin{matrix}\%m_{Al}=\dfrac{5,4}{6,6}×100\%=81,82\%\\\%m_{Mg}=\dfrac{1,2}{6,6}×100\%=18,18\%\end{matrix}\right.$
