\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 2Al+3Cl_2\underrightarrow{^{to}}2AlCl_3\\ n_{Cl_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ \Rightarrow V=V_{Cl_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\ \Rightarrow m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)