-calculez masa de brom din solutie
c/100= md/ms; 4/100= md/200---> md= 8gBr2 initial
2,04/100=md/200---> md= 3,92g dupa reactie
-calculez moli Br2 intrati in reactie
n=(8-3,92)g/160g/mol= 0,0245mol Br2
-ecuatia reactiei:
1mol 1mol
CnH2n +Br2---> CnH2nBr2
n...........0,0245
n=0,0245mol CnH2n
n=mxM---> 0,0245=1,4xM--> M aproximativ 56g/mol
-deduc alchena 14n=56--> n= 4=>C4H8
CH2=C(CH3)-CH3 2metil propena