\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.1......0.3............0.1.........0.15\)
\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)
\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)
\(C\%_{HCl}=\dfrac{10.95}{100}\cdot100\%=10.95\%\)
\(m_{dd}=2.7+100-0.15\cdot2=102.4\left(g\right)\)
\(m_{AlCl_3}=0.1\cdot133.5=13.35\left(g\right)\)
\(C\%_{AlCl_3}=\dfrac{13.35}{102.4}\cdot100\%=13.04\%\)