\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.1........0.3.........0.1...........0.15\)
\(m_{dd_{HCl}}=97.8\cdot1=97.8\left(g\right)\)
\(m_{ddsaupư}=2.7+97.8-0.15\cdot2=100.2\left(g\right)\)
\(C\%_{AlCl_3}=\dfrac{0.1\cdot133.5}{100.2}\cdot100\%=13.32\%\)