Em tham khảo nha:

Em xem lại đề câu 1 nhé 

\(\begin{array}{l}
1)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
4Na + {O_2} \xrightarrow{t^0} 2N{a_2}O\\
hh:Al(a\,mol),Na(b\,mol)\\
{n_{A{l_2}{O_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,5a\,mol\\
{n_{N{a_2}O}} = \dfrac{{{n_{Na}}}}{2} = 0,5b\,mol\\
\left\{ \begin{array}{l}
27a + 23b = 20\\
0,5a \times 102 + 0,5b \times 62b = 22,8
\end{array} \right.\\
 \Rightarrow a = b = \\
2)\\
4Al + 3{O_2} \xrightarrow{t^0}2A{l_2}{O_3}\\
2Cu + {O_2} \xrightarrow{t^0} 2CuO\\
hh:Al(a\,mol),Cu(b\,mol)\\
{n_{{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
\left\{ \begin{array}{l}
27a + 64b = 11,8\\
0,75a + 0,5b = 0,2
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,1\\
{m_{Al}} = 0,2 \times 27 = 5,4g\\
{m_{Cu}} = 11,8 - 5,4 = 6,4g
\end{array}\)