$n_{Al}=\dfrac{2,7}{27}=0,1\ (mol)$
$2Al+6HCl\to 2AlCl_3+3H_2\\0,1\quad\ \ \ \xrightarrow{\qquad}\ \ \quad 0,1\ \to \ 0,15\ \ (mol)$
$\Rightarrow \begin{cases}m_{H_2}=0,15.2=0,3\ (g)\\m_{AlCl_3}=0,1.133,5=13,35\ (g)\end{cases}$
$m_{dd\ spứ}=m_{Al}+m_{dd\ HCl}-m_{H_2}$
$\Rightarrow m_{dd\ spứ}=2,7+50-0,3=52,4\ (g)$
$C\%_{dd\ AlCl_3}=\dfrac{13,35}{52,4}.100\%=25,47\%$