a) PTPU:
`2Al + 6HCl -> 2AlCl_3 + 3H_2`
`n_(Al)=(5,4)/(27)=0,2` (mol)
Theo PTHH, ta có: `n_(H_2)=3/2n_(Al)=3/2 . 0,2=0,3` mol
`-> V_(H_2)=0,3.22,4=6,72` (l)
b) `n_(AlCl_3)=n_(Al)=0,2` mol
`-> m_(AlCl_3)=0,2.133,5=26,7` (g)
c) `n_(HCl)=3n_(Al)=0,2.3=0,6` mol
`-> V_(HCl)=(0,6)/(0,8)=0,75` (l)