Đáp án:

\(\begin{array}{l}
a)\\
\% Fe = 41,2\% \\
\% F{e_2}{O_3} = 58,8\% \\
b)\\
C{\% _{FeC{l_2}}} = 11,2\% \\
C{\% _{FeC{l_3}}} = 14,33\% \\

\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{m_{HCl}} = \dfrac{{100 \times 18,25}}{{100}} = 18,25g\\
{n_{HCl}} = \dfrac{{18,25}}{{36,5}} = 0,5mol\\
hh:Fe(a\,mol),F{e_2}{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
56a + 160b = 13,6\\
2a + 6b = 0,5
\end{array} \right.\\
 \Rightarrow a = 0,1mol;b = 0,05mol\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% Fe = \dfrac{{5,6}}{{13,6}} \times 100\%  = 41,2\% \\
\% F{e_2}{O_3} = 100 - 41,2 = 58,8\% \\
b)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
{m_{FeC{l_2}}} = 0,1 \times 127 = 12,7g\\
{n_{FeC{l_3}}} = 2F{e_2}{O_3} = 0,1mol\\
{m_{FeC{l_3}}} = 0,1 \times 162,5 = 16,25g\\
{n_{{H_2}}} = {n_{Fe}} = 0,1mol\\
{m_{{\rm{dd}}spu}} = 13,6 + 100 - 0,1 \times 2 = 133,4g\\
C{\% _{FeC{l_2}}} = \dfrac{{12,7}}{{113,4}} \times 100\%  = 11,2\% \\
C{\% _{FeC{l_3}}} = \dfrac{{16,25}}{{113,4}} \times 100\%  = 14,33\% \\

\end{array}\)