initial : msi = 40g c= 31,5% ⇒ mdi = 12,6g
final : msf = 40 - 4 - 5,3 = 30,7g
1mol cristalohidrat ...........(A+106,5)g sare anhidra................108g H2O
5,3g.............................................x
x = 5,3(A+106,5)/(A + 214,5)
mdf = 12,6 - x
(12,6-x)·100/30,7 = 31,5
1260 - 100x = 967,05 ⇒ x = 2,9295
2,9295(A+214,5) = 5,3(A+ 106,5) A + 214,5 = 1,81A + 192,678
0,81A = 21,82 A ≈ 27 g/moli ⇒ M = Al⁺³
