Bảo toàn e:
$2n_{SO_2}=n_{FeO}$
$\to n_{SO_2}=0,1(mol)$
Gọi chung hai bazơ là $ROH$ ($0,07+0,06=0,13$ mol)
$\overline{M}_{ROH}=\dfrac{0,07.56+0,06.40}{0,13}=\dfrac{632}{13}$
$\to \overline{M}_R=\dfrac{411}{13}$
$\dfrac{n_{ROH}}{n_{SO_2}}=1,3$
$\to$ tạo 2 muối: $RHSO_3$ ($x$), $R_2SO_3$ ($y$)
$\to \begin{cases} x+y=0,1\\ x+2y=0,13\end{cases}$
$\to \begin{cases} x=0,07\\ y=0,03\end{cases}$
$\to m=0,07.\left(\dfrac{411}{13}+81\right)+0,03\left(\dfrac{411}{13}.2+80\right)=12,18g$