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