$m_{CuO}=20\%.50=10g$
$⇒n_{CuO}=10/80=0,125mol$
$⇒m_{Fe_2O_3}=50-10=40g$
$⇒n_{Fe_2O_3}=40/160=0,25mol$
$a.CuO+H_2\overset{t^o}\to Cu+H_2O(1)$
$ Fe_2O_3+3H_2\overset{t^o}\to 2Fe+3H_2O(2)$
$b.n_{H_2}=n_{H_2(1)}+n_{H_2(2)}=0,125+0,75=0,875mol$
$⇒V_{H_2}=0,875.22,4=19,6l$
c.Theo pt (1) :
$n_{Cu}=n_{CuO}=0,125mol$
$⇒m_{Cu}=0,125.64=8g$
Theo pt (2) :
$n_{Fe}=2.n_{Fe_2O_3}=2.0,25=0,5mol$
$⇒m_{Fe}=0,25.56=28g$
$\%m_{Cu}=\dfrac{8}{8+28}.100\%=22,22\%$
$\%m_{Fe}=100\%-22,22\%=77,78\%$