\(a.n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ 0,1...........0,3...........0,1.........0,15\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ b.C_{MddHCl}=\dfrac{0,3}{0,1}=3\left(M\right)\\ c.m_{AlCl_3}=133,5.0,1=13,35\left(g\right)\)
