Đáp án:
a) 20,88 g
b) 77,31 g
Giải thích các bước giải:
a)
$\begin{gathered}
2FeO + 4{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + S{O_2} + 4{H_2}O \hfill \\
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 \\
2Fe{(OH)_2} + 4{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + S{O_2} + 6{H_2}O \hfill \\
2FeC{O_3} + 4{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + S{O_2} + 2C{O_2} + 4{H_2}O \hfill \\
\end{gathered} $
${n_{S{O_2}}} = \dfrac{{1,344}}{{22,4}} = 0,06mol$
Theo PTHH: ${n_X} = 2{n_{S{O_2}}} = 0,12mol$
$ \Rightarrow {n_{F{e_3}{O_4}}} = {n_X}.25\% = 0,03mol$
$\begin{gathered}
\Rightarrow {n_{{\text{c}}{\text{.ran con lai}}}} = 0,12 - 0,03 = 0,09 \hfill \\
\Rightarrow \sum {{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{3}{2}{n_{F{e_3}{O_4}}} + } \dfrac{1}{2}{n_{{\text{c}}{\text{.ran con lai}}}} = 0,09mol \hfill \\
\Rightarrow {m_{F{e_2}{{(S{O_4})}_3}}} = 0,09.400 = 36g \hfill \\
\Rightarrow m + 15,12 = 36 \Rightarrow m = 20,88g \hfill \\
\end{gathered} $
b)
$\begin{gathered}
F{e_2}{(S{O_4})_3} + 3Ba{(OH)_2} \to 3BaS{O_4} \downarrow + 2Fe{(OH)_3} \downarrow \hfill \\
2Fe{(OH)_3}\xrightarrow{{{t^o}}}F{e_2}{O_3} + 3{H_2}O \hfill \\
\end{gathered} $
Chất rắn sau khi nung là $F{e_2}{O_3}$, $BaS{O_4}$
$\begin{gathered}
{n_{F{e_2}{O_3}}} = {n_{F{e_2}{{(S{O_4})}_3}}} = 0,09mol \hfill \\
{n_{BaS{O_4}}} = 3{n_{F{e_2}{{(S{O_4})}_3}}} = 0,27mol \hfill \\
\Rightarrow a = {m_{F{e_2}{O_3}}} + {m_{BaS{O_4}}} = 0,09.160 + 0,27.233 = 77,31g \hfill \\
\end{gathered} $
