\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Gọi CT của oxit Fe là: FexOy
\(Fe_xO_y+yH_2\rightarrow xFe+yH_2O\)
7,2 g -----------------5,6g
n (mol)---------------0,1(mol)
\(\Rightarrow n_{Fe_xO_y}=\dfrac{0,1.7,2}{5,6}\approx0,1\left(mol\right)\)
\(M_{Fe_xO_y}=\dfrac{7,2}{0,1}=72\) (g/mol)
Ta có: Fex + Oy = 72
56x + 16y = 72
\(\Rightarrow x=y=1\)
\(\Rightarrow CT:FeO\)