Giải thích các bước giải: 

\(\begin{array}{l}
a)F{e_3}{O_4} + 8HCl \to FeC{l_2} + 2FeC{l_3} + 4{H_2}O\\
b)\\
{n_{F{e_3}{O_4}}} = 0,08mol\\
{m_{HCl}} = \dfrac{{14,6\%  \times 200}}{{100\% }} = 29,2g\\
 \to {n_{HCl}} = 0,8mol\\
 \to {n_{HCl}} > {n_{F{e_3}{O_4}}} \to {n_{HCl}}dư\\
 \to {n_{FeC{l_2}}} = {n_{F{e_3}{O_4}}} = 0,08mol\\
 \to {n_{FeC{l_3}}} = 2{n_{F{e_3}{O_4}}} = 0,16mol\\
 \to {m_{FeC{l_2}}} = 10,16g\\
 \to {m_{FeC{l_3}}} = 26g\\
{m_{{\rm{dd}}}} = {m_{F{e_3}{O_4}}} + {m_{{\rm{dd}}HCl}} = 18,56 + 200 = 218,56g\\
 \to C{\% _{FeC{l_2}}} = \dfrac{{10,16}}{{218,56}} \times 100\%  = 4,65\% \\
 \to C{\% _{FeC{l_3}}} = \dfrac{{26}}{{218,56}} \times 100\%  = 11,9\% 
\end{array}\)

 a)nZn =13/65=.2

     nS= 4.8/32 

Zn      +      S     ->    ZnS

     0.2      .15           .15      (mol)

b)mZnS= .15*97=14.55

c)  ZnS + HCl -> ZnCl + H2S

   .15                               .15  mol

V H2S = .15 * 22.4 = 3.36