\(n_{HCl\left(bđ\right)}=2\left(mol\right);n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Vì:\dfrac{2}{6}>\dfrac{0,05}{3}\Rightarrow HCldư\\ a,n_{Al}=\dfrac{2}{3}.0,05=\dfrac{1}{30}\left(mol\right)\\ m_{Al}=\dfrac{1}{30}.27=0,9\left(g\right)\\ b,n_{HCl\left(dư\right)}=2-0,05.2=1,9\left(mol\right)\)