Đáp án:

\(\begin{array}{l}
a)\\
\% {m_{Cu}} = 65,31\% \\
\% {m_{Mg}} = 16,33\% \\
\% {m_{Al}} = 18,36\% \\
b)\\
{C_\% }HCl = 1,63\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
hh:Cu(a\,mol),Al(b\,mol),Mg(c\,mol)\\
{n_{{H_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5\,mol\\
\left\{ \begin{array}{l}
64a + 27b + 24c = 29,4\\
1,5b + c = 0,5\\
64a = 19,2
\end{array} \right.\\
 \Rightarrow a = 0,3;b = c = 0,2\\
\% {m_{Cu}} = \dfrac{{0,3 \times 64}}{{29,4}} \times 100\%  = 65,31\% \\
\% {m_{Mg}} = \dfrac{{0,2 \times 24}}{{29,4}} \times 100\%  = 16,33\% \\
\% {m_{Al}} = 100 - 65,31 - 16,33 = 18,36\% \\
b)\\
AlC{l_3} + 3AgN{O_3} \to 3AgCl + Al{(N{O_3})_3}\\
MgC{l_2} + 2AgN{O_3} \to Mg{(N{O_3})_2} + 2AgCl\\
HCl + AgN{O_3} \to AgCl + HN{O_3}\\
{n_{AgCl}} = \dfrac{{200,9}}{{143,5}} = 1,4\,mol\\
{n_{HCl}}  \text{ dư }= 1,4 - 0,2 \times 3 - 0,2 \times 2 = 0,4\,mol\\
{n_{HCl}} \text{ ban đầu }= 0,5 \times 2 + 0,4 = 1,4\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{1,4}}{{0,5}} = 2,8l \Rightarrow {m_{{\rm{dd}}HCl}} = 1,12 \times 2800 = 3136g\\
{C_\% }HCl = \dfrac{{1,4 \times 36,5}}{{3136}} \times 100\%  = 1,63\% 
\end{array}\)