$m_M=29,6-12,8=16,8g$

$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$

$2M+2nHCl\to 2MCl_n+ nH_2$

$\Rightarrow n_M=\dfrac{0,6}{n}(mol)$

$M_M=\dfrac{16,8n}{0,6}=28n$

$n=2\Rightarrow M_M=56(Fe)$

$n_{Cu}=\dfrac{12,8}{64}=0,2(mol)$

$CuO+CO\to CO_2+Cu$

$\Rightarrow n_{CuO}=0,2(mol)$

$\Rightarrow m_{Fe_xO_y}=39,2-0.2.80=23,2g$

$Fe_xO_y+yCO\to xFe+yCO_2$

$m_{Fe}=16,8g\Rightarrow n_{Fe}=\dfrac{16,8}{56}=0,3(mol)$

$\Rightarrow n_{Fe_xO_y}=\dfrac{0,3}{x}(mol)$

$M_{Fe_xO_y}=\dfrac{23,2x}{0,3}=\dfrac{232x}{3}=56x+16y$

$\Leftrightarrow \dfrac{x}{y}=\dfrac{3}{4}$

$\Rightarrow x=3; y=4$

Vậy oxit $M_xO_y$ là $Fe_3O_4$