Question

Exercitiul 14 va rog mult

Answer 1

MOLI NaOH -total:
c=n/V; 1= n/0,08---> n=0,08molNaOH
MOLI NaOH -exces=MOLI HCl
2=n/0,01---> n=0,02mol NaOH
MOLI NaOH  REACTIONATI CU TRIGLICERIDA =0,06mol
1 mol triglicerida.....3molNaOH
n=0,02......................0,06
n=m/M       0,02= 16,12/M---> M,triglicerida = 806g/mol
formula generala : C6H5O6R3,unde R este radical hidrocarbonat
M ,C6H5O6=173g---> M,3R= 806-173=633g
                                       M,R=  633/3g= 211g
daca R este saturat, are formula CnH2n+1--.> 14n =210---n=15--> acidul palmitic--->1,2,3 tripalmitil glicerol