\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right);n_{HCl}=\dfrac{25,55}{36,5}=0,7\left(mol\right)\\ Vì:\dfrac{0,2}{2}< \dfrac{0,7}{6}\Rightarrow HCldư\\ n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\ m=m_{muối}=m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)