$m_{CuO}=25\%.16=4g⇒n_{CuO}=4/80=0,05mol$
$⇒m_{Fe_2O_3}=16-4=12g⇒n_{Fe_2O_3}=12/160=0,075mol$
$PTHH :$
$CuO+H_2\overset{t^o}\to Cu+H_2O$
$Fe_2O_3+3H_2\overset{t^o}\to 2Fe+3H_2O$
$a/n_{Fe}=2.n_{Fe_2O_3}=2.0,075=0,15mol⇒m_{Fe}=0,15.56=8,4g$
$n_{Cu}=n_{CuO}=0,05mol⇒m_{Cu}=0,05.64=3,2g$
$b/V_{H_2}=(0,05+0,225).22,4=6,16l$