$m_{CuO}=30,3.26,4\%=8(g)$

$\Rightarrow m_{PbO}=30,3-8=24,3(g)$

$\Rightarrow n_{CuO}=\dfrac{8}{80}=0,1(mol)$

$n_{PbO}=\dfrac{22,3}{223}=0,1(mol)$

$PTHH:CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$PbO+H_2\xrightarrow{t^o}Pb+H_2O$

Theo PT: $n_{Cu}=n_{CuO}=0,1(mol);n_{Pb}=n_{PbO}=0,1(mol)$

$\Rightarrow m_{Cu}=0,1.64=6,4(g);m_{Pb}=0,1.207=20,7(g)$

Đáp án:

\(\begin{array}{l}
{m_{Cu}} = 6,4g\\
{m_{Pb}} = 20,7g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
PbO + {H_2} \xrightarrow{t^0} Pb + {H_2}O\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
{m_{CuO}} = 30,3 \times 26,4\%  = 8g\\
{m_{PbO}} = 30,3 - 8 = 22,3g\\
{n_{CuO}} = \dfrac{8}{{80}} = 0,1\,mol \Rightarrow {n_{Cu}} = {n_{CuO}} = 0,1\,mol\\
{n_{PbO}} = \dfrac{{22,3}}{{223}} = 0,1\,mol \Rightarrow {n_{Pb}} = {n_{PbO}} = 0,1\,mol\\
{m_{Cu}} = 0,1 \times 64 = 6,4g\\
{m_{Pb}} = 0,1 \times 207 = 20,7g
\end{array}\)