$pthh :$
$CuO+H2→Cu+H2O$
$Fe2O3+3H2→2Fe+3H2O$
Gọi $n_{CuO}=x (mol) ; n_{Fe2O3}=y(mol)$
theo pt :
$n_{H_{2}}=n_{H_{2}(1)}+n_{H_{2}(2)}=x+3y=0,7mol$
$m_{Cu}+m_{Fe}=64x+112y=28,8g$
Ta có hpt :
$\left \{ {{x+3y=0,7} \atop {64x+112y=28,8}} \right.$
Giải hệ ta đc :
$\left \{ {{x=0,1} \atop {y=0,2}} \right.$
$⇒m=m_{CuO}+m_{Fe2O3}=80x+160y=80.0,1+160.0,2=40g$
