$\text{Cho số mol của PbO và} Fe_3O_4 \text{là x,y (mol)}$
$n_{H_2}=\dfrac{17,92}{22,4}=0,8(mol)$
$\text{a)PTHH:}$
$PbO+H_2\xrightarrow{t^0}Pb+H_2O$
x __ x __ x __ x $\text{(mol)}$
$Fe_3O_4+4H_2\xrightarrow{t^0}3Fe+4H_2O$
y ___ 4y ___ 3y __ 4y $\text{(mol)}$
$\text{Ta có:}$
$x+4y=0,8(1)$
$m_{Pb}+m_{Fe}=66,5$
$\to 207x+168y=66,5(2)$
$\text{Từ (1),(2) suy ra:}$
$x=0,2(mol);y=0,15(mol)$
$\text{b)}$
$m_{\text{..hỗn hợp oxit}}=m_{PbO}+m_{Fe_3O_4}=44,6+34,8=79,4g$
%$m_{PbO}=\dfrac{\text{223.0,2.100%}}{79,4}=56,17$%
%$m_{Fe_3O_4}=\dfrac{\text{232.0,15.100%}}{79,4}=43,8$%