Răspuns:
doar pentena reactioneaza cu bromul
CH2=CH-CH2-CH2-CH3 + Br2---> CH2Br-CHBr-CH2-CH2-CH3
in raport molar 1:1
c,m=niu/V---> niu= 2mol/lx0,060 l= 0,12mol Br2
deci in amestec se afla 0,12molC5H10,
M,C5H10=70g/mol---> m=0,12x70g= 8,4g pentena
in amestec se afla: 10g-8,4g=1,6g pentan
10g amestec.......1,6g pentan.......8,4g pentena
100g.....................................x..............................y
x= 16% pentan
y=84% pentena
Explicație: