Giải thích các bước giải:
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{n_{{H_2}}} = 0,1mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
\to {m_{Fe}} = 0,1 \times 56 = 5,6g\\
\to {m_{F{e_2}{O_3}}} = 13,6 - 5,6 = 8g\\
\to {n_{F{e_2}{O_3}}} = 0,05mol\\
\to \% {m_{Fe}} = \dfrac{{5,6}}{{13,6}} \times 100\% = 41,18\% \\
\to \% {m_{F{e_2}{O_3}}} = \dfrac{8}{{13,6}} \times 100\% = 58,82\% \\
a)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,1mol\\
{n_{HCl(pt)}} = 2{n_{Fe}} + 6{n_{F{e_2}{O_3}}} = 0,5mol\\
{n_{HCl(lt)}} = 0,6mol \to {n_{HCl(dư)}} = 0,1mol\\
\to C{M_{FeC{l_3}}} = \dfrac{{0,05}}{{0,6}} = 0,083M\\
\to C{M_{FeC{l_2}}} = C{M_{HCl}} = \dfrac{{0,1}}{{0,6}} = 0,167M\\
b)\\
2NaCl + {H_2}S{O_4} \to 2HCl + N{a_2}S{O_4}\\
{n_{NaCl}} = {n_{HCl}} = 0,6mol\\
\to {m_{NaCl(lt)}} = 35,1g\\
\to {m_{NaCl(tt)}} = \dfrac{{35,1 \times 75\% }}{{100\% }} = 26,33g
\end{array}\)