de la 1 la 15:
100% ..................... 66,66 g Al impur
81% ....................... m.pura = 54 g Al
54g m g n moli
2Al + 3Cl2 --> 2AlCl3
2x27 3x71 2
=> m = 54x3x71/2x27 = 213 g Cl2 consumat
=> n = 54x2/2x27 = 2 moli AlCl3