Hóa trị của Fe: 2y/x
nHCl = 0,8 . 0,6 = 0,48 mol
Pt: FexOy + 2yHCl --> xFeCl2y/x + yH2O
.....\(\dfrac{0,24}{y}\)<---0,48
Ta có: \(13,92=\dfrac{0,24}{y}.\left(56x+16y\right)\)
\(\Leftrightarrow13,92=\dfrac{13,44x}{y}+3,84\)
\(\Leftrightarrow13,44x=10,08y\)
\(\Leftrightarrow\dfrac{x}{y}=\dfrac{10,08}{13,44}=\dfrac{3}{4}\)
Vậy CTHH: Fe3O4