a,
$m_{CuO}= 16.25\%= 4g$
$\Rightarrow n_{CuO}= \frac{4}{80}= 0,05 mol$
$m_{FeO}= 16-4=12g$
$\Rightarrow n_{FeO}= \frac{12}{72}= \frac{1}{6} mol$
$CuO+ H_2 \buildrel{{t^o}}\over\to Cu+ H_2O$
$FeO+ H_2 \buildrel{{t^o}}\over\to Fe+ H_2O$
$n_{Cu}= 0,05 mol \Rightarrow m_{Cu}= 0,05.64= 3,2g$
$n_{Fe}= \frac{1}{6} mol \Rightarrow m_{Fe}= \frac{1}{6}.56= 9,33g$
b,
$n_{H_2}= 0,05+\frac{1}{6}= \frac{13}{60} mol$
$\Rightarrow V_{H_2}= 22,4.\frac{13}{60}= 4,85l$