a) %O = 9·16·100/250 = 57,6%
b) initial : ms = 275 + 125 = 400g md= 80g
deoarece in 125 g CuSO5·5H2O se gasesc 125·160/250 = 80g CuSO4
(250g CuSO4·5H2O.......................160g CuSO4
125.....................................................x )
c = 80·100/ 400 = 20%