`n_{H_2} = (14,874)/(24,79) = 0,6` mol

`n_{Al} = (0,54)/(27) = 0,02` mol

Nếu `HCl` dư

`2Na + 2HCl -> 2NaCl + H_2(1)`

`2Al + 6HCl -> 2AlCl_3 + 3H_2(2)`

Theo PTHH:

`n_{HCl} =2n_{H_2(1)} + 3n_{Al} = 2 . 0,6 + 3 . 0,02 = 1,26` mol

`=> a = (1,26)/(0,5) = 2,52(M)`

Theo PT:

`n_{Na} = 2n_{H_2] = 2 . 0,6 = 1,2` mol

`=> m = 1,2 . 23 = 27,6(g)`

Nếu `Na` dư

`2Na + 2HCl -> 2NaCl + H_2(1)`

`2Na + 2H_2O -> 2NaOH + H_2(2)`

`2Al + 2H_2O + 2NaOH -> 2NaAlO_2 + 3H_2(3)`

Theo PTHH `(3)`

`n_{NaOH(dư)} = n_{Al} = 0,02` mol

Theo PTHH `(2)`

`n_{H_2(2)} = 1/2n_{NaOH} = 1/2 . 0,02 = 0,01` mol

`=> n_{H_2(1)} = 0,6 - 0,01 = 0,59` mol

Theo PTHH `(1):`

`n_{HCl} = 2n_{H_2(1)} = 2 . 0,59 = 1,18` mol

`=> a = (1,18)/(0,5) = 2,36(M)`

Theo PTHH:

`n_{Na} = n_{NaOH} + 2n_{H_2(1)} = 0,02 + 2 . 0,59 = 1,2` mol

`=> m = 1,2 . 23 = 27,6(g)`