\(M_xO_y+2yHCl\rightarrow xMCl_{\frac{2y}{x}}+yH_2O\)
0,8/y_______________________0,8
\(\Rightarrow M_{oxit}=\frac{46,4}{0,8}=58y=M_x+16y\)
\(\Rightarrow M_x=42y\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\Rightarrow M=56\left(Fe\right)\)
Vậy oxit là Fe3O4