Giải thích các bước giải:

\(\begin{array}{l}
Fe + S \to FeS\\
FeS + {H_2}S{O_4} \to FeS{O_4} + {H_2}S\\
{n_{Fe}} = 0,16mol\\
{n_S} = 0,15mol\\
 \to {n_{Fe}} > {n_S} \to {n_{Fe}}dư\\
 \to {n_{FeS}} = {n_S} = 0,15mol\\
 \to {n_{FeS{O_4}}} = 0,15mol\\
A.\\
2FeS + 10{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 10{H_2}O + 9S{O_2}\\
{n_{S{O_2}}} = \dfrac{9}{2}{n_{FeS}} = 0,675mol\\
{n_{Ba{{(OH)}_2}}} = 0,26mol\\
 \to \dfrac{{{n_{Ba{{(OH)}_2}}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,26}}{{0,675}} = 0,385
\end{array}\)

-> Tạo 1 muối : \(B{\rm{aS}}{O_3}\)

\(\begin{array}{l}
S{O_2} + Ba{(OH)_2} \to B{\rm{aS}}{O_3} + {H_2}O\\
{n_{{\rm{BaS}}{O_3}}} = {n_{S{O_2}}} = 0,675mol\\
 \to {m_{{\rm{BaS}}{O_3}}} = 0,675 \times 217 = 146,5g\\
B.\\
6FeS{O_4} + {K_2}C{r_2}{O_7} + 7{H_2}S{O_4} \to 3F{e_2}{(S{O_4})_3} + 7{H_2}O + {K_2}S{O_4} + C{r_2}{(S{O_4})_3}\\
{n_{{K_2}C{r_2}{O_7}}} = \dfrac{1}{6}{n_{FeS{O_4}}} = 0,025mol\\
 \to {m_{{K_2}C{r_2}{O_7}}} = 0,025 \times 294 = 7,35g
\end{array}\)