Đáp án:
\(\begin{array}{l}
\% Mg = 47,06\% \\
\% Al = 52,94\% \\
m = 22,85g\\
C{\% _{HCl}} = 9,125\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
hh:Mg(a\,mol),Al(b\,mol)\\
a + \dfrac{3}{2}b = 0,25(1)\\
24a + 27b = 5,1(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,1;b = 0,1\\
{m_{Mg}} = 0,1 \times 24 = 2,4g\\
{m_{Al}} = 5,1 - 2,4 = 2,7g\\
\% Mg = \dfrac{{2,4}}{{5,1}} \times 100\% = 47,06\% \\
\% Al = 100 - 47,06 = 52,94\% \\
{n_{AlC{l_3}}} = {n_{Al}} = 0,1mol\\
{m_{AlC{l_3}}} = 0,1 \times 133,5 = 13,35g\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,1mol\\
{m_{MgC{l_2}}} = 0,1 \times 95 = 9,5g\\
m = 13,35 + 9,5 = 22,85g\\
{n_{HCl}} = 2{n_{Mg}} + 3{n_{Al}} = 0,5mol\\
{m_{HCl}} = 0,5 \times 36,5 = 18,25g\\
C{\% _{HCl}} = \dfrac{{18,25}}{{200}} \times 100\% = 9,125\%
\end{array}\)
