Quy đổi hỗn hợp gồm $Fe$ (x mol), $O$ (y mol)
$\Rightarrow 56x+16y=22,4$ $(1)$
Bán phản ứng:
$\mathop{Fe}\limits^0 \to \mathop{Fe}\limits^{+3}+3e$
$\mathop{O}\limits^0+2e\to \mathop{O}\limits^{-2}$
$\mathop{S}\limits^{+6}+2e\to \mathop{S}\limits^{+4}$
Bảo toàn e: $3n_{Fe}=2n_O+2n_{SO_2}$
$\Rightarrow n_{SO_2}=\dfrac{3x-2y}{2}= 1,5x- y(mol)$
Bảo toàn $Fe$: $n_{Fe_2(SO_4)_3}=\dfrac{n_{Fe}}{2}=0,5x (mol)$
Bảo toàn $S$:
$n_{H_2SO_4}=3n_{Fe_2(SO_4)_3}+n_{SO_2}$
$\Rightarrow 3.0,5x+1,5x-y=0,55$
$\Leftrightarrow 3x-y=0,55$ $(2)$
Từ $(1)(2)\Rightarrow x=0,3; y=0,35$
$n_{Fe_2(SO_4)_3}=0,5x=0,15(mol)$
$\to m_{\text{muối}}=0,15.400=60g$