Giải thích các bước giải:
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}(1)\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O(2)\\
{m_{HCl}} = \dfrac{{91,25 \times 20\% }}{{100\% }} = 18,25g\\
\to {n_{HCl}} = 0,5mol
\end{array}\)
Gọi a và b lần lượt là số mol của Fe và Fe2O3
\(\begin{array}{l}
\left\{ \begin{array}{l}
56a + 160b = 13,6g\\
2a + 6b = 0,5
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,1\\
b = 0,05
\end{array} \right.\\
\to {n_{Fe}} = 0,1mol \to {m_{Fe}} = 0,1 \times 56 = 5,6g\\
\to \% {m_{Fe}} = \dfrac{{5,6}}{{13,6}} \times 100\% = 41,18\% \\
\to {n_{F{e_2}{O_3}}} = 0,05mol \to {m_{F{e_2}{O_3}}} = 0,05 \times 160 = 8g\\
\to \% {m_{F{e_2}{O_3}}} = \dfrac{8}{{13,6}} \times 100\% = 58,82\% \\
b)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1mol\\
\to {m_{FeC{l_2}}} = 0,1 \times 127 = 12,7g\\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,1mol\\
\to {m_{FeC{l_3}}} = 0,1 \times 162,5 = 16,25g\\
\end{array}\)
\(\begin{array}{l}
\to {m_{{\rm{ddFeC}}{{\rm{l}}_2}}} = {m_{{\rm{ddFeC}}{{\rm{l}}_3}}} = {m_{honhop}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}}\\
\to {m_{{\rm{ddFeC}}{{\rm{l}}_2}}} = {m_{{\rm{ddFeC}}{{\rm{l}}_3}}} = 13,6 + 91,25 - 0,1 \times 2 = 104,65g\\
\to C{\% _{{\rm{ddFeC}}{{\rm{l}}_2}}} = \dfrac{{12,7 \times 100\% }}{{104,65}} = 12,14\% \\
\to C{\% _{{\rm{ddFeC}}{{\rm{l}}_3}}} = \dfrac{{16,25 \times 100\% }}{{104,65}} = 15,53\%
\end{array}\)
