Em tham khảo nha :

\(\begin{array}{l}
a)\\
CuO + {H_2} \to Cu + {H_2}O\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
hh:CuO(a\,mol);F{e_2}{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
80a + 160b = 40\\
64a + 112b = 28,8
\end{array} \right.\\
 \Rightarrow a = 0,1mol;b = 0,2mol\\
{m_{CuO}} = 0,1 \times 80 = 8g\\
\% CuO = \dfrac{8}{{40}} \times 100\%  = 20\% \\
\% F{e_2}{O_3} = 100 - 20 = 80\% \\
b)\\
{n_{{H_2}}} = {n_{CuO}} + 3{n_{F{e_2}{O_3}}} = 0,7mol\\
{V_{{H_2}}} = 0,7 \times 22,4 = 15,68l\\
c)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = \dfrac{7}{{15}}mol\\
{m_{Al}} = \dfrac{7}{{15}} \times 27 = 12,6g
\end{array}\)

Cho `CuO: x(mol), Fe_2O_3: y(mol)`

`=> 80x+160y=40g(1)`

`CuO+H_2\overset{t^o}{\to}Cu+H_2O`

`Fe_2O_3+3H_2\overset{t^o}{\to}2Fe+3H_2O`

`=> 64x+112y=28,8g(2)`

`(1),(2)=>x=0,1(mol), y=0,2(mol)`

`a)` `%m_{CuO}=\frac{0,1.80.100%}{40}=20%`

`=> %m_{Fe_2O_3}=100%-20%=80%`

`b)` `n_{H_2}=x+3y=0,7(mol)`

`=> V_{H_2}=0,7.22,4=15,68(l)`

`c)` `2Al+6HCl\to 2AlCl_3+3H_2`

`=> n_{Al}=\frac{2}{3}.0,7=\frac{7}{15}(mol)`

`=> m_{Al}=\frac{7}{15}.27=12,6g`