C4H10 nu reactioneaza cu Br2 ori a moli
C4H8 + Br2---> C4H8Br2 ori b moli
1mol.......1mol
calculez moli brom
c= n/V
1= n/0,020---> n= 0,02mol= b mol butena
sumarul maselor:
3,44= ax58g butan+ bx56gbutena
inlocuind b, rezulta a= 0,04mol butan
n,butan: n,butena= 0,04:0,02=........................