$n_{Fe}=\dfrac{12,6}{56}=0,225(mol)$
$n_{SO_2}=\dfrac{4,2}{22,4}=0,1875(mol)$
Gọi chung hh $X$ là $Fe_xO_y$.
$2Fe_xO_y+(6x-2y)H_2SO_4\to xFe_2(SO_4)_3+(3x-2y)SO_2+(6x-2y)H_2O$
$n_{Fe_xO_y}=\dfrac{n_{Fe}}{x}=\dfrac{0,225}{x}(mol)$
$\to \dfrac{0,225}{2x}.(3x-2y)=0,1875$
$\to 0,1125(3x-2y)=0,1875x$
$\to 0,15x=0,225y$
$\to y=\dfrac{2}{3}x$
Suy ra CTTQ hỗn hợp là $Fe_xO_{\frac{2}{3}x}$
$\to m=\dfrac{0,225}{x}.\Big(56x+\dfrac{2}{3}.16x\Big)=15g$
