Question

14 g amestec echimolar 1-butena si 3-hexena se oxideaza cu o solutie de K2Cr2O7/H2SO4 0,5M.Substanta organica obtinuta se dizolva in 177,8 g apa.Concentratia procentuala a solutiei de substanta organica obtinuta in final este:A.11,1% B.22,2% C.1,11% D.2,22% E.7,6%

Answer 1

Răspuns:

M,butena=56g/mol

M,HEXENA= 84g/mol

CH2=CH-CH2-CH3+5O====>CO2+H2O+ CH3-CH2-COOH    ori  a  moli

CH3-CH2-CH=CH-CH2-CH3 +4O===>2CH3-CH2-COOH............ori   a  moli

total masa hidrocarburi>

ax(56+84)  =14g=====> a=0,1mol

total moli si masa acid propanoic..

niu=3x a,mol       m= niuxM=0,3molx74g/mol=22,2gaceasta este masa dizolvata in apa, notata md

ms= md+m,apa= 22,2+177,8=> 200g solutie

c= mdx100/ms= 22,2x100/200%=11,1%

Explicație: