\(n_{Fe_xO_y}=\dfrac{23,2}{56x+16y}\left(mol\right)\)
PTHH: 2FexOy + (6x-2y)H2SO4(đ) -to-> xFe2(SO4)3 + (3x-2y)SO2 + (6x-2y)H2O
=> \(n_{Fe_2\left(SO_4\right)_3}=\dfrac{11,6x}{56x+16y}\left(mol\right)\)
=> \(m_{Fe_2\left(SO_4\right)_3}=\dfrac{11,6x}{56x+16y}.400=60\)
=> 11,6x = 8,4x + 2,4y
=> x : y = 3 : 4
=> CTHH: Fe3O4