Question

O masa de solutie de 42,4g amestec pentan si pentena decoloreaza 320g Br2 de concentratie 20%. Care e raportul molar pentan:pentena?
Raspunsul e 1:3, dar mie imi da 1:2.

Answer 1

Cp=md/ms*100==>md=Cp*ms/100=20*320/100=64g Br2

n=m/M=64/160=0.4 moli Br2

CH3-CH2-CH2-CH2-CH3 + Br2 ≠

1mol...................................................................1mol
CH2=CH-CH2-CH2-CH3 + Br-Br ---> CH2(Br)-CH(Br)-CH2-CH2-CH3
0.4mol...............................................................0.4mol

n2=4 moli ==> m1=70*0.4=28g C5H10

m2=42.4-28=14.4g C5H12

n1=m/M=14.4/72=0.2 moli C5H12

n1/n2=0.2/0.4=2/4=1/2


MC5H10=70g/mol
MC5H12=72g/mol
MBr2=160g/mol