Question

56 grame amestec de acid acetic si acid malonic se obtin prin oxidarea 2,5 heptadienei. Volumul solutiei semimolare de KMnO4/H2SO4 consumat este egal cu....

Answer 1

CH3-CH=CH-CH2-CH=CH-CH3 + 8O--. >2CH3COOH+HOOC-CH2-COOH

M,C7H14= 98g/mol

M,acid acetic= 60g/mol

M,acid malonic= 104g/mol

din ecutia chimica reazulta:

din 98gheptadiena...........(2x60+104)g acizi

x............................................56g

x=24,5g

niu, heptadiena= m/M=0,25mol----> niuO= 8X0,25 mol

2mol...................................................................................5mol

2KMno4+ 3H2SO4 --->K2SO4+ 2MnSO4 + 3H2O+5O

x.................................................................................................2mol

x= 0,8mol KMnO4

C= 0,5mol/l

C= niu/V   --------> V=0,8/0,5=1,6 L