CTHH: \(Fe_xO_y\)
\(n_{CO}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(n_{Fe_xO_y}=\frac{14,4}{56x+16y}\left(mol\right)\)
PTHH: \(Fe_xO_y+yCO\underrightarrow{t^o}xFe+yCO_2\)
________\(\frac{0,2}{y}\) <-----0,2__________________(mol)
=> x = y
=> FeO