Giải thích các bước giải:
\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = 0,065mol
\end{array}\)
Gọi a và b lần lượt là số mol của Al và Fe
\(\begin{array}{l}
\left\{ \begin{array}{l}
27a + 56b = 1,93\\
\dfrac{3}{2}a + b = 0,065
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,03\\
b = 0,02
\end{array} \right.\\
\to {n_{Al}} = 0,03mol\\
\to {n_{Fe}} = 0,02mol\\
a)\\
\% {m_{Al}} = \dfrac{{0,03 \times 27}}{{1,93}} \times 100\% = 41,97\% \\
\% {m_{Fe}} = 100\% - 41,97\% = 58,03\% \\
b)\\
{n_{HCl}} = 3{n_{Al}} + 2{n_{Fe}} = 0,13mol\\
\to {m_{HCl}} = 4,745g\\
\to {m_{{\rm{dd}}HCl}} = \dfrac{{4,745}}{{18\% }} \times 100\% = 26,36g\\
c)\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,03mol \to {m_{AlC{l_3}}} = 4,005g\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,02mol \to {m_{FeC{l_2}}} = 2,54g\\
{m_{{\rm{dd}}Y}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 28,16g\\
\to C{\% _{AlC{l_3}}} = \dfrac{{4,005}}{{28,16}} \times 100\% = 14,22\% \\
\to C{\% _{FeC{l_2}}} = \dfrac{{2,54}}{{28,16}} \times 100\% = 9\% \\
d)\\
{V_{HCl}} = \dfrac{{26,36}}{{1,1}} = 23,96ml = 0,024l\\
\to C{M_{AlC{l_3}}} = \dfrac{{0,03}}{{0,024}} = 1,25M\\
\to C{M_{FeC{l_2}}} = \dfrac{{0,02}}{{0,024}} = 0,83M
\end{array}\)
