\(n_{Al}=\dfrac{0,054}{27}=0,002\left(mol\right)\\ n_{HCl}=0,02.0,4=0,008\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ LTL:\dfrac{0,002}{2}< \dfrac{0,008}{6}\Rightarrow HCldư\\ n_{AlCl_3}=n_{Al}=0,002\left(mol\right)\\ \Rightarrow m_{AlCl_3}=0,002.133,5=0,267\left(g\right)\\ n_{HCl\left(dư\right)}=0,008-0,002.3=0,002\left(mol\right)\\ \Rightarrow m_{HCl}=0,0002.36,5=0,073\left(g\right)\)
=> Chọn A