Em tham khảo nha:

\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
m = {m_{hh}} - {m_{{H_2}}} \Rightarrow {m_{{H_2}}} = 7,8 - 7 = 0,8g\\
{n_{{H_2}}} = \dfrac{{0,8}}{2} = 0,4\,mol\\
hh:Al(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
1,5a + b = 0,4\\
27a + 24b = 7,8
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,1\\
{m_{Mg}} = 0,1 \times 24 = 2,4g\\
{m_{Al}} = 7,8 - 2,4 = 5,4g\\
\% {m_{Mg}} = \dfrac{{2,4}}{{7,8}} \times 100\%  = 30,77\% \\
\% {m_{Al}} = 100 - 30,77 = 69,23\% \\
b)\\
{n_{HCl}} \text{ phản ứng } = 2{n_{{H_2}}} = 0,8\,mol\\
HCl + KOH \to KCl + {H_2}O\\
{n_{HCl}}  \text{ dư }= {n_{KOH}} = 0,1 \times 0,02 = 0,002\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{(0,8 + 0,002) \times 36,5}}{{20\% }} = 146,365g
\end{array}\)