Question

66 va rog :). Va rog este urgent . Mulțumesc

Answer 1

daca triglicerida ar contine 2 resturi de acid palmitic si 1 de acid oleic:
acid palmitic CH3-(CH2)14- COOH      M = 256g mol
acid oleic : CH3-(CH2)7-CH=CH-(CH2)7-COOH    M = 282 g/mol
glicerina : C3H8O3
---------------------------------
M triglicerida = 2·256 + 282 + 92 - 3·18 = 832g/mol ≠858
daca triglicerida = dioleopalmitina : M = 2·282 +256 + 92 - 54 = 858 g/mol 
triglicerida + 3NaOH ⇒ CH3-(CH2)14-COONa + 2C17H33-COONa + C3H8O3
nNaOH necesar = 3 moli  n folosit = 3+50/100 ·3 = 4,5 moli  m= 4,5·40 = 180g
ms = 180·100/30 = 600g
initial in solutia de NaOH + NaCl :  msi = 600 + 200 = 800g;
m sapun = 2·304 + 278 = 886 g
msf = 800 - 120 + 886 +92 = 1658g
% sapun = 886·100/ 1658 = 53,437%