Em tham khảo nha :
\(\begin{array}{l}
a)\\
CuO + {H_2} \to Cu + {H_2}O\\
PbO + {H_2} \to Pb + {H_2}O\\
{n_{{H_2}O}} = \dfrac{m}{M} = \dfrac{{5,4}}{{18}} = 0,3mol\\
hh:CuO(a\,mol),Pb(b\,mol)\\
a + b = 0,3(1)\\
80a + 223b = 38,3(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,2mol;b = 0,1mol\\
{m_{CuO}} = n \times M = 0,2 \times 80 = 16g\\
\% CuO = \dfrac{{16}}{{38,3}} \times 100\% = 41,8\% \\
\% PbO = 100 - 41,8 = 58,2\% \\
b)\\
{n_{Cu}} = {n_{CuO}} = 0,2mol\\
{m_{CuO}} = n \times M = 0,2 \times 64 = 12,8g\\
{n_{Pb}} = {n_{PbO}} = 0,1mol\\
{m_{Pb}} = n \times M = 0,1 \times 207 = 20,7g\\
c)\\
{n_{{H_2}}} = {n_{{H_2}O}} = 0,3mol\\
{V_{{H_2}}} = n \times 22,4 = 0,3 \times 22,4 = 6,72l
\end{array}\)