$n_{H_2O}=1,26/18=0,07mol$
$CuO+H_2\overset{t^o}\to Cu+H_2O$
$PbO+H_2\overset{t^o}\to Pb+H_2O$
$\text{Gọi}$ $n_{CuO}=a;n_{PbO}=b$
$\text{Ta có :}$
$m_{hh}=80a+223b=5,64$
$n_{H_2O}=a+b=0,07$
$\text{Ta có hpt :}$
$\left\{\begin{matrix}
80a+223b=5,64 & \\
a+b=0,07 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=0,07 & \\
b=2.10^{-4} &
\end{matrix}\right.$
$⇒\%m_{CuO}=\dfrac{0,06.80.100\%}{5,64}=85,1\%$
$\%m_{PbO}=100\%-85,1\%=14,9\%$