Đáp án:
b) $8,4(l)$
c) $50g$
Giải thích các bước giải:
a)
$\begin{gathered}
3Fe + 2{O_2} \to F{e_3}{O_4} \hfill \\
4Fe + 3{O_2} \to 2F{e_2}{O_3} \hfill \\
2Fe + {O_2} \to 2FeO \hfill \\
\end{gathered} $
$\begin{gathered}
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 \hfill \\
2FeO + 4{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + S{O_2} + 4{H_2}O \hfill \\
\end{gathered} $
$F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O$
$2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O$
b)
${n_{Fe}} = \dfrac{{14}}{{56}} = 0,25mol$
Bảo toàn electron: $2{n_{S{O_2}}} = 3{n_{Fe}} = 0,75 \Rightarrow {n_{S{O_2}}} = 0,375mol$
$ \Rightarrow {V_{S{O_2}}} = 0,375.22,4 = 8,4(l)$
c)
Bảo toàn nguyên tố $Fe$: ${n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Fe}} = 0,125mol$
$ \Rightarrow {m_{F{e_2}{{(S{O_4})}_3}}} = 0,125.400 = 50g$
