$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$
$Fe_xO_y+ yH_2\to xFe+yH_2O$
$\Rightarrow n_{Fe_xO_y}=\dfrac{0,3}{x}(mol)$
$n_{CuO}=\dfrac{8}{80}=0,1(mol)$
$CuO+H_2\to Cu+H_2O$
$2Cu+O_2\to 2CuO$
$\Rightarrow n_{CuO\text{bđ}}=0,1(mol)$
Ta có:
$\dfrac{0,3}{x}(56x+16y)+0,1.80=31,2$
$\Leftrightarrow 56x+16y=\dfrac{232x}{3}$
$\Leftrightarrow \dfrac{x}{y}=\dfrac{3}{4}$
$\Rightarrow x=3; y=4$
Vậy oxit sắt có CTHH là $Fe_3O_4$