$m_{CuO}=50.20\%=10g$
$\Rightarrow n_{CuO}=\dfrac{10}{80}=0,125 mol$
$n_{FeO}=\dfrac{50-10}{72}=\dfrac{5}{9}mol$
$CuO+H_2\to Cu+H_2O$
$FeO+H_2\to Fe+H_2O$
$\Rightarrow n_{H_2}=0,125+\dfrac{5}{9}=\dfrac{49}{72}mol$
$\to V_{H_2}=22,4.\dfrac{49}{72}=15,24l$