Gọi CTHH oxit là $M_2O_n$
$n_M=\dfrac{16,8}{M}(mol)$
$n_{M_2O_n}=\dfrac{23,2}{2M+16n}(mol)$
$4M+nO_2\to 2M_2O_n$
$\Rightarrow 2n_{M_2O_n}=n_M$
$\Rightarrow \dfrac{46,4}{2M+16n}=\dfrac{16,8}{M}$
$\Rightarrow 16,8(2M+16n)=46,4M$
$\Rightarrow M=21n$
$\to n=\dfrac{8}{3}; M=56(Fe)$
Vậy kim loại là sắt, oxit là $Fe_2O_{\frac{8}{3}}$ hay $Fe_3O_4$