Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Mg}} = 30,77\% \\
\% {m_{Al}} = 69,23\% \\
b)\\
a = 1,6
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
hh:Mg(a\,mol),Al(b\,mol)\\
\left\{ \begin{array}{l}
a + 1,5b = 0,4\\
24a + 27b = 7,8
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
\% {m_{Mg}} = \dfrac{{0,1 \times 24}}{{7,8}} \times 100\% = 30,77\% \\
\% {m_{Al}} = 100 - 30,77 = 69,23\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,8\,mol\\
{C_M}HCl = \dfrac{{0,8}}{{0,5}} = 1,6M \Rightarrow a = 1,6
\end{array}\)
