Đáp án:
\(\begin{array}{l}
a)\\
\% Al = 24,3\% \\
\% Fe = 75,7\% \\
b)\\
C{\% _{AlC{l_3}}} = 8,32\% \\
C{\% _{FeC{l_2}}} = 11,87\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3mol\\
hh:Al(a\,mol);Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 11,1\\
\dfrac{3}{2}a + b = 0,3
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,15\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
\% Al = \dfrac{{2,7}}{{11,1}} \times 100\% = 24,3\% \\
\% Fe = 100 - 24,3 = 75,7\% \\
b)\\
{n_{HCl}} = 3{n_{Al}} + 2{n_{Fe}} = 0,6mol\\
{m_{HCl}} = 0,6 \times 36,5 = 21,9g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{21,9 \times 100}}{{14,6}} = 150g\\
{m_{{\rm{dd}}spu}} = 150 + 11,1 - 0,3 \times 2 = 160,5g\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1mol\\
{m_{AlC{l_3}}} = 0,1 \times 133,5 = 13,35g\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,15mol\\
{m_{FeC{l_2}}} = 0,15 \times 127 = 19,05g\\
C{\% _{AlC{l_3}}} = \dfrac{{13,35}}{{160,5}} \times 100\% = 8,32\% \\
C{\% _{FeC{l_2}}} = \dfrac{{19,05}}{{160,5}} \times 100\% = 11,87\%
\end{array}\)
