Ta có: $\rm n_{H_2}=\dfrac{16,8}{22,4}=0,75\ (mol)$
$\rm\quad\quad\;\;\;2Al+6HCl\to 2AlCl_3+3H_2\\(mol)\;\;0,5\hspace{,7cm}\;1,5\hspace{1,cm}\;\;\;0,5\leftarrow\hspace{,0cm}0,75$
Từ phản ứng, suy ra:
$\rm m_{Al}=0,5\times 27=13,5\ (gam)$
$\rm m_{AlCl_3}=0,5\times 133,5=66,75\ (gam)$
$\rm C_{M\ HCl}=\dfrac{1,5}{1}=1,5\ (M)$