a,
Gọi $x$, $y$ là số mol $Fe$, $Fe_3O_4$
$\Rightarrow 56x+232y=17,2$ $(1)$
$n_{SO_2}=\dfrac{3,92}{22,4}=0,175(mol)$
$2Fe+6H_2SO_4\to Fe_2(SO_4)_3+3SO_2+6H_2O$
$2Fe_3O_4+10H_2SO_4\to 3Fe_2(SO_4)_3+SO_2+10H_2O$
$\Rightarrow \dfrac{3}{2}x+\dfrac{1}{2}y=0,175$ $(2)$
$(1)(2)\Rightarrow x=0,1; y=0,05$
$\%m_{Fe}=\dfrac{0,1.56.100}{17,2}=32,56\%$
b,
$n_{Ca(OH)_2}=0,1.1,5=0,15(mol)$
$SO_2+Ca(OH)_2\to CaSO_3+H_2O$
Ban đầu tạo $0,15$ mol $CaSO_3$. Còn dư $0,175-0,15=0,025$ mol $SO_2$
$SO_2+CaSO_3+H_2O\to Ca(HSO_3)_2$
Tạo ra $0,025$ mol $Ca(HSO_3)_2$. Dư $0,15-0,025=0,125$ mol $CaSO_3$
$\to m_{\text{muối}}=0,025.202+0,125.120=20,05g$
