Question

O grasime are masa moleculara 838g peste o cantitate din aceasta se adaoga in exces 1,2l solutie KOH de concentratie 3M.Iar excesul este neutralizat de 110g solutie HCl de concentratie 36,5%.Ce cantitate de grasime a fost folosita initial?

Answer 1

Calculam cantitatea de KOH introdusa in proces
Cm=n/Vs => n=Cm*Vs=3*1,2=3,6moli KOH
Calculam excesul de KOH
c/100=md/ms => md=c*ms/100=36,5*110/100=40,15g HCl
1mol..36,5g
KOH + HCl -> KCl + H2O
x=?....40,15g
x=40,15/36,5=1,1moli KOH
nKOH necesari=3,6-1,1=2,5moli
838g..........3moli
Grasime + 3KOH -> 3sapun de K + glicerina
y=?............2,5moli
y=838*2,5/3=698,33g grasime