Question

3.44g amestec butan butena decoloreaza 20 ml solutie de brom in tetraclorura de carbon . Raportul molar butena -butan este:




(24)

Answer 1

Cm = niu/Vs, Cm = 1 M

Vs = 20 mL = 0,02 L

niu = CmxVs = 1x0,02 = 0,02 moli Br

m g      0,02 moli

C4H8 + Br2 --CCl4--> C4H8Br2

 56          1

=> m = 56x0,02/1 = 1,12 g C4H8

=> 3,44 - 1,12 = 2,32 g C4H10

1,12/56 = 0,02 moli C4H8

2,32/58 = 0,04 moli C4H10

=> total 0,06 moli amestec

=> %C4H8 = 0,02x100/0,06 = 33,33%

=> %C4H10 = 100 - 33,33 = 66,67%