Giải:
a, PTHH: `2Al + 6HCl -> 2AlCl_3 + 3H_2`
b, `n_{H_2} = (3,36)/(22,4) = 0,15` (mol)
Theo PTHH, `n_{Al} = 1,5n_{H_2} = 1,5 . 0,15 = 0,225` (mol)
c, Theo PTHH, `n_{HCl} = 2n_{H_2} = 2 . 0,15 = 0,3` (mol)
d, Theo PTHH, `n_{AlCl_3} = n_{Al} = 0,225` (mol)
`m_{AlCl_3} = 0,225 . (27 + 35,5 . 3) = 30,0375` (g)