Đặt CT oxit là $R_xO_y$
$n_{SO_2}=\dfrac{3,248}{22,4}=0,145(mol)$
$x\mathop{R}\limits^{+\frac{2y}{x}}\to x\mathop{R}\limits^{+n}+(nx-2y)e$
$\mathop{S}\limits^{+6}+2e\to \mathop{S}\limits^{+4}$
Bảo toàn e:
$2.0,145=\dfrac{nx-2y}{x}.n_R$
Mà $n_R=x.n_{R_xO_y}$
$\to (nx-2y)n_{R_xO_y}=0,29$
$\to (nx-2y).\dfrac{20,88}{xM_R+16y}=0,29$
$\to nx-2y=\dfrac{xM_R+16y}{72}$
$\to M_R=\dfrac{72(nx-2y)-16y}{x}=\dfrac{72nx-16y-144}{x}$
$\to x=y=1; n=3; M_R=56(Fe)$
$\to FeO$