Em tham khảo nha:

\(\begin{array}{l}
5)\\
1:\,N{a_2}O\\
2:Pb\\
PbO + {H_2} \xrightarrow{t^0} Pb + {H_2}O\\
3:\,A{l_2}{O_3}\\
4:Fe\\
F{e_2}{O_3} + 3{H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
5:\,Cu\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
6:MgO\\
6)\\
a)\\
F{e_2}{O_3} + 3{H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_{F{e_2}{O_3}}} = \dfrac{{0,3}}{3} = 0,1\,mol\\
{m_{F{e_2}{O_3}}} = 0,1 \times 160 = 16g\\
{m_{A{l_2}{O_3}}} = 26,2 - 16 = 10,2g\\
b)\\
{n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,2\,mol\\
m = {m_{Fe}} + {m_{A{l_2}{O_3}}} = 0,2 \times 56 + 10,2 = 21,4g\\
c)\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
{n_{CuO}} = \dfrac{{16}}{{80}} = 0,2\,mol\\
{n_{{H_2}}} > {n_{CuO}} \Rightarrow {H_2} \text{ dư }\\
{n_{Cu}} = {n_{CuO}} = 0,2\,mol\\
{m_{Cu}} = 0,2 \times 64 = 12,8g
\end{array}\)