Giải thích các bước giải:

\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Mg + 2{H_2}S{O_4} \to Mg{\rm{S}}{O_4} + S{O_2} + 2{H_2}O(1)\\
Cu + 2{H_2}S{O_4} \to C{\rm{uS}}{O_4} + S{O_2} + 2{H_2}O(2)
\end{array}\)

\(\begin{array}{l}
a)\\
{m_{Cu}} = 1,28g \to {n_{Cu}} = 0,02mol\\
{n_{{H_2}}} = 0,04mol\\
{n_{S{O_2}}} = 0,06mol\\
 \to {n_{S{O_2}(2)}} = {n_{Cu}} = 0,02mol\\
 \to {n_{S{O_2}(1)}} = 0,04mol\\
 \to {n_{Mg}} = {n_{S{O_2}}} = 0,04mol\\
 \to {m_{Mg}} = 0,04 \times 24 = 0,96g\\
 \to {m_{Al}} = 4,12 - (0,96 + 1,28) = 1,88g \to {n_{Ag}} = 0,02mol\\
 \to \% {m_{Al}} = \dfrac{{1,88}}{{4,12}} \times 100\%  = 45,63\% \\
 \to \% {m_{Mg}} = \dfrac{{0,96}}{{4,12}} \times 100\%  = 23,3\% \\
 \to \% {m_{Cu}} = \dfrac{{1,28}}{{4,12}} \times 100\%  = 31,07\% \\
b)\\
{n_{HCl}} = 3{n_{Al}} + 2{n_{Mg}} = 0,14mol
\end{array}\)

\(\begin{array}{l}
 \to {m_{HCl}} = 0,14 \times 36,5 = 5,11g\\
 \to {m_{{\rm{dd}}HCl}} = (\dfrac{{5,11}}{{7,3\% }} \times 100\% ) + 10\%  = 70,1g\\
{n_{AlC{l_3}}} = {n_{Ag}} = 0,02mol \to {m_{AlC{l_3}}} = 0,02 \times 133,5 = 2,67g\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,04mol \to {m_{MgC{l_2}}} = 0,04 \times 95 = 3,8g\\
 \to {m_{{\rm{dd}}Y}} = {m_{KL}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 4,12 + 70,1 - 0,04 \times 2 = 74,14g\\
 \to C{\% _{AlC{l_3}}} = \dfrac{{2,67}}{{74,14}} \times 100\%  = 3,6\% \\
 \to C{\% _{MgC{l_2}}} = \dfrac{{3,8}}{{74,14}} \times 100\%  = 5,13\% 
\end{array}\)