$n_{H_2}=\dfrac{0,672}{22,4}=0,03 mol$
$Fe+2HCl\to FeCl_2+H_2$
$\Rightarrow n_{Fe}=0,03 mol$
$m_{Fe}=0,03.56=1,68g$
$\Rightarrow n_{Cu}=\dfrac{2,64-1,68}{64}=0,015 mol$
$CuO+H_2\to Cu+H_2O$
$\Rightarrow n_{CuO}=n_{Cu}=0,015 mol= n_{Fe_xO_y}$
$m_{CuO}=0,015.80=1,2g$
$\Rightarrow m_{Fe_xO_y}=3,36-1,2=2,16g$
$M_{Fe_xO_y}=\dfrac{2,16}{0,015}=144$ (bạn xem lại đề)