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Tham khảo :

`-` Quá trình :

$X+HNO_3 \to Fe(NO_3)_3+NO+H_2SO_4+H_2O \xrightarrow{+Ba(OH)_2}Fe(OH)_3+BaSO_4$

`-` Quy về :

$\begin{cases} Fe:a\\S:b \end{cases}$

`->`$56a+32b=20,8$$(1)$

`-` Bảo toàn nguyên tố $Fe:$

$n_{Fe}=n_{Fe(OH)_3}=a$

`-` Bảo toàn nguyên tố $S:$

$n_{S}=n_{BaSO_4}=b$

`->`$107a+233b=91,3$$(2)$

`-` Từ $(1),(2)$ suy ra : $a=0,2;b=0,3$

`-` Bảo toàn $e:$

$3n_{Fe}+6n_{S}=n_{NO}$

`=>`$n_{NO}=3 \cdot 0,2 + 6 \cdot 0,3=2,4(mol)$

`=>`$V=53,76(l)$

$\rm x \,\, 1|$$16H_2O+Fe+FeS+FeS_2+S \to 4SO_4^{2-}+3Fe^{3+}+33e^-+32H^+$

$\rm x \,\, 33|$$2H^++NO_3^- +e^- \to NO_2+H_2O$

`->`$Fe+FeS+FeS_2+S+34H^++33NO_3^{-} \to 3Fe^{3+}+33NO_2+4SO_4^{2-}+17H_2O$

`=>`$n_{H^+}≈n_{NO_2}=2,4(mol)$