\(n_{Al}=\dfrac{10.8}{27}=0.4\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.4........1.2.........0.4..........0.6\)
\(m_{HCl}=1.2\cdot36.5=43.8\left(g\right)\)
\(m_{AlCl_3}=0.4\cdot133.5=53.4\left(g\right)\)
\(m_{dd}=10.8+100-0.6\cdot2=109.6\left(g\right)\)
\(C\%_{AlCl_3}=\dfrac{53.4}{109.6}\cdot100\%=48.72\%\)