$\text{Cho số mol của PbO và} Fe_3O_4 \text{là x,y (mol)}$

$n_{H_2}=\dfrac{17,92}{22,4}=0,8(mol)$

$\text{a)PTHH:}$

$PbO+H_2\xrightarrow{t^0}Pb+H_2O$

x  __   x  __   x    __  x $\text{(mol)}$

$Fe_3O_4+4H_2\xrightarrow{t^0}3Fe+4H_2O$

y  ___   4y  ___ 3y  __  4y   $\text{(mol)}$

$\text{Ta có:}$

$x+4y=0,8(1)$

$m_{Pb}+m_{Fe}=66,5$

$\to 207x+168y=66,5(2)$

$\text{Từ (1),(2) suy ra:}$

$x=0,2(mol);y=0,15(mol)$

$\text{b)}$

$m_{\text{..hỗn hợp oxit}}=m_{PbO}+m_{Fe_3O_4}=44,6+34,8=79,4g$

%$m_{PbO}=\dfrac{\text{223.0,2.100%}}{79,4}=56,17$%

%$m_{Fe_3O_4}=\dfrac{\text{232.0,15.100%}}{79,4}=43,8$%