a)
$\begin{align}
& 2Fe+{{O}_{2}}\xrightarrow{t{}^\circ }2FeO \\
& 4Fe+2{{O}_{2}}\xrightarrow{t{}^\circ }2F{{e}_{2}}{{O}_{3}} \\
& 3Fe+2{{O}_{2}}\xrightarrow{t{}^\circ }F{{e}_{3}}{{O}_{4}} \\
\end{align}$
Hỗn hợp A gồm: Fe dư, $FeO,F{{e}_{2}}{{O}_{3}},F{{e}_{3}}{{O}_{4}}$
$\begin{align}
& Fe+6{{H}_{2}}S{{O}_{4}}\to F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+3S{{O}_{2}}+6{{H}_{2}}O \\
& 2FeO+4{{H}_{2}}S{{O}_{4}}\to F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+S{{O}_{2}}+4{{H}_{2}}O \\
& 2F{{e}_{3}}{{O}_{4}}+10{{H}_{2}}S{{O}_{4}}\to 3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+S{{O}_{2}}+10{{H}_{2}}O \\
\end{align}$
b)
Bảo toàn khối lượng:
${{m}_{Fe}}+{{m}_{O}}={{m}_{A}}\Rightarrow {{m}_{O}}={{m}_{A}}-{{m}_{Fe}}=17,4-12,6=4,8g$
${{n}_{O}}=\frac{4,8}{16}=0,3mol$
${{n}_{Fe}}=\frac{12,6}{56}=0,225mol$
Bảo toàn electron, ta có:
$3{{n}_{Fe}}=2{{n}_{O}}+2{{n}_{S{{O}_{2}}}}\Rightarrow {{n}_{S{{O}_{2}}}}=\frac{3{{n}_{Fe}}-2{{n}_{O}}}{2}=\frac{3\times 0,225-2\times 0,3}{2}=0,0375mol$
${{V}_{S{{O}_{2}}}}=0,0375\times 22,4=0,84(l)$
c)
Bảo toàn nguyên tố S, ta có:
${{n}_{{{H}_{2}}S{{O}_{4}}}}=3{{n}_{F{{e}_{2}}{{(S{{O}_{4}})}_{3}}}}+{{n}_{S{{O}_{2}}}}$
${{n}_{F{{e}_{2}}{{(S{{O}_{4}})}_{3}}}}=\frac{1}{2}{{n}_{Fe}}=\frac{1}{2}\times 0,225=0,1125mol$
${{n}_{{{H}_{2}}S{{O}_{4}}}}=3\times 0,1125+0,0375=0,375mol$
${{m}_{{{H}_{2}}S{{O}_{4}}}}=0,375\times 98=36,75g$
${{m}_{\text{dd}{{H}_{2}}S{{O}_{4}}}}=\frac{36,75\times 100}{98}=37,5g$
${{V}_{{{H}_{2}}S{{O}_{4}}}}=37,5:1,84=20,38(ml)$