M.CaBr2 = 40+80x2 = 200 g/mol
20g m g
Ca + Br2 --> CaBr2
40 200
=> m = 20x200/40 = 100 g CaBr2 obtinuti teoretic (Ct) la 100%, dar la 80% randament cat s-ar obtine.....?
100% ...................... 100g CaBr2
80% ....................... Cp = 80g CaBr2