Gọi CTHH là FexOy
\(n_{CO}=0,4\left(mol\right)\)
\(Fe_xO_y+yCO\underrightarrow{^{to}}xFe+yCO_2\)
0.4/y_____0.4 _____________
\(m_{FexOy}=\frac{0,4}{y\left(56x+16y\right)}=23,2\)
\(\Rightarrow m=22,4x+6,4y=23,2y\)
\(\Rightarrow22,4x=16,8y\)
\(\Leftrightarrow\frac{x}{y}=\frac{3}{4}\)
Vậy CTHH là Fe3O4