Giải thích các bước giải:
Tự luận:
\(\begin{array}{l}
1.\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
{m_{HCl}} = \dfrac{{100 \times 36,5\% }}{{100\% }} = 36,5g\\
\to {n_{HCl}} = 1mol\\
\to {n_{HCl(1)}} = 2{n_{{H_2}}} = 0,3mol\\
\to {n_{HCl(2)}} = 0,7mol\\
\to {n_{AlC{l_3}}} = {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,1mol\\
\to {m_{AlC{l_3}}} = 0,1 \times 133,5 = 13,35\\
\to {m_{Al}} = 0,1 \times 27 = 2,7g\\
{n_{FeC{l_2}}} = {n_{FeO}} = \dfrac{1}{2}{n_{HCl(2)}} = 0,35mol\\
\to {m_{FeC{l_2}}} = 0,35 \times 127 = 44,45g\\
\to {m_{FeO}} = 0,35 \times 72 = 25,2g\\
\to {m_{{\rm{dd}}}} = {m_{hỗnhợp}} + {m_{HCl}} - {m_{{H_2}}} = (2,7 + 25,2) + 100 - 0,15 \times 2\\
\to {m_{{\rm{dd}}}} = 127,6g\\
\to C{\% _{AlC{l_3}}} = \dfrac{{13,35}}{{127,6}} \times 100\% = 10,46\% \\
\to C{\% _{FeC{l_2}}} = \dfrac{{44,45}}{{127,6}} \times 100\% = 34,84\% \\
2.\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
{n_{{H_2}}} = 0,2mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,2mol\\
\to {m_{Fe}} = 0,2 \times 56 = 11,2g\\
\to {m_{A{l_2}{O_3}}} = 21,4 - 11,2 = 10,2g\\
\to {n_{A{l_2}{O_3}}} = 0,1mol\\
\to {n_{HCl}} = {n_{Fe}} + 6{n_{A{l_2}{O_3}}} = 0,8mol\\
\to {m_{HCl}} = 0,8 \times 36,5 = 29,2g\\
\to {m_{{\rm{dd}}HCl}} = \dfrac{{29,2}}{{20\% }} \times 100\% = 146g\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,2mol\\
{n_{AlC{l_3}}} = 2{n_{A{l_2}{O_3}}} = 0,2mol\\
\to {m_{FeC{l_2}}} = 0,2 \times 127 = 25,4g\\
\to {m_{AlC{l_3}}} = 0,2 \times 133,5 = 26,7g\\
\to {m_{{\rm{dd}}}} = {m_{hỗnhợp}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 21,4 + 146 - 0,4 = 167g\\
\to C{\% _{FeC{l_2}}} = \dfrac{{25,4}}{{167}} \times 100\% = 15,2\% \\
\to C{\% _{AlC{l_3}}} = \dfrac{{26,7}}{{167}} \times 100\% = 16\%
\end{array}\)
