AgBr = 188 g/mol
188 g AgBr ........................... 80 g
94 g AgBr ........................... m = 40 g Br
CnH2n+2 + Br2 --lumina--> CnH2n+1Br + HBr
14n+2 14n+81
36 g alcan + 40 g Br = 76 g produs bromurat
76 g ........................... 40 g Br
14n+81 ..................... 80 g Br
=> n = 5 , B)