$a) PTPƯ: 2Al + 6HCl → 2AlCl_{3} + 3H_{2}↑$ (1)
-Theo bài ra, ta có:
$n_{Al} = \dfrac{5,4}{27} = 0,2$ $(mol)$
-Theo (1): $n_{AlCl_{3}} = n_{Al} = 0,2$ $mol$
$⇒ m_{AlCl_{3}} = 0,2.133,5 = 26,7$ $(g)$
b) -Theo (1): $n_{H_{2}} = \dfrac{3}{2}.n_{Al} = \dfrac{3}{2}.0,2 = 0,3$ $(mol)$
$⇒ V_{H_{2_{đktc}}} = 0,3.22,4 = 6,72$ $(l)$