Đáp án:

\(\begin{array}{l}
a)\quad \%m_{Mg} =57,14\%\\
\qquad\%m_{Al} = 42,86\%\\
b)\quad m_{muối} = 55,2\ g\\
c)\quad V_{HCl} = 210,24\ \rm ml
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
\rm Mg \quad + \quad 2HCl \quad\longrightarrow\quad MgCl_2\quad+\quad H_2\uparrow\\
a\ \text{mol}\qquad 2a\ \text{mol}\qquad \qquad a\ \text{mol}\qquad \qquad a\ \text{mol}\\
\rm 2Al \quad + \quad 6HCl \quad\longrightarrow\quad 2AlCl_3\quad+\quad 3H_2\uparrow\\
b\ \text{mol}\qquad 3b\ \text{mol}\qquad \qquad b\ \text{mol}\qquad \qquad \dfrac32b\ \text{mol}\\
n_{H_2} = \dfrac{13,44}{22,4} = 0,6\ \rm mol\\
a)\\
\text{Hỗn hợp:}\quad a\ \text{mol  Mg},\ b \ \text{mol  Al}\\
\Rightarrow \begin{cases}2a + 27b = 12,6\\a + \dfrac32b = 0,6\end{cases} \Leftrightarrow \begin{cases}a = 0,3\\b=0,2\end{cases}\\
\%m_{Mg} = \dfrac{0,3\times 24\times 100\%}{12,6}=57,14\%\\
\%m_{Al} = 100\% - 57,14\% = 42,86\%\\
b)\\
n_{MgCl_2} = n_{Mg} = 0,3\ \rm mol\\
n_{AlCl_3} = n_{Al} = 0,2\ \rm mol\\
m_{muối} = m_{MgCl_2} + m_{AlCl_3} = 0,3\times95 + 0,2\times 133,5 = 55,2\ g\\
c)\\
n_{HCl\ pứ} = 2n_{Mg} + 3n_{Al} = 0,3\times 2 + 0,2\times 3 = 1,2\ \rm mol\\
\Rightarrow V_{HCl\ pứ} = \dfrac{\times m}{d\times C\%} = \dfrac{1,2\times 36,5}{1,25\times 0,2} = 175,2\ \rm ml\\
\Rightarrow V_{HCl\ lấy} = V_{HCl\ pứ}\times 120\% = 210,24\ \rm ml
\end{array}\)