Theo đề bài, ta có hệ phương trình:
$\left \{ {{m_{Fe}+m_{Cu}=29,6} \atop {m_{Fe}-m_{Cu}=4}} \right.⇒\left \{ {{m_{Fe}=16,8\ g} \atop {m_{Cu}=12,8\ g}} \right.$
$n_{Fe}=16,8:56=0,3mol$
$n_{Cu}=12,8:64=0,2mol$
$PTHH:CuO+H_2--t^°->Cu+H_2O$ $(1)$
$FeO+H_2--t^°->Fe+H_2O$ $(2)$
Theo $PTHH(1):n_{H_2\ (1)}=n_{Cu}=0,2mol$
$(2):n_{H_2\ (2)}=n_{Fe}=0,3mol$
Tổng $n_{H_2}=0,2+0,3=0,5mol$
$V_{H_2}=0,5.22,4=11,2l$
$n_{CuO}=n_{Cu}=0,2mol$
$n_{FeO}=n_{Fe}=0,3mol$
$→m=0,2.80+0,3.72=37,6g$