Đặt CTHH chung là `Fe_xO_y`
Ta có: `M_{Fe}=56x`
`M_{O}=16y`
`=>` $\frac{M_{Fe}}{M_O}$ = $\frac{56x}{16y}$
mà $\frac{M_{Fe}}{M_O}$ = $\frac{70}{30}$ = $\frac{7}{3}$
`=>` $\frac{56x}{16y}$ = $\frac{7}{3}$
`<=>` `56x.3=16y.7`
`<=>` `168x=112y`
`<=>` $\frac{x}{y}$ = $\frac{112}{168y}$ = $\frac{2}{3}$
`=>` `CTHH:Fe_2O_3` (sắt oxide)
b,
A là oxit bazơ
`2FeO + 4H_2SO_4→ Fe_2(SO_4)_3 + SO_2 + 4H_2O`