I need help help with #2 on the first image and I think the second image is used

Question

I need help help with #2 on the first image and I think the second image is used to figure it out 1. Use the volume of HCl from the titrations to calculate the moles of OH in each flask. [3 pts] Blank Butter moles of OH in each flask Soy oil 2.20 mol Show your calculations here: ? mol 0H--?nel oc -H t : Cx V H CL 0.250.60 390 mol 0.230.20 nla a 3mmo 6-2 50+5.50 : 2. 20 mo/ 2. Calculate the moles of OH-consumed by the saponification reactions [2 pts] Soy oil Butter moles of OH consumed by TAG Show your calculations here: # 1 Consume d-CTorgha m6GTT(1ewah h )

Explanation / Answer

General saponification reaction is given by:

CH3-(CH2)nCOO-CH2-CH(OOC(CH2)n-CH3)-CH2-OOC(CH2)n-CH3 (triglyceride ,oil or fat) +3NaOH --->

3 Na-OOC(CH2)n-CH3 + HO-CH2-CH(OH)-CH2OH (glycerol)

So ,NaOH, in excess , istaken and oil added to undergo saponification,some amount of NaOH is used up in saponification ,and remaining is neutralized by HCl (back titration)

HCl +NaOH --->NaCl+H2O

So, mol HCl /molNaOH=1:1 .They react in 1:1 molar ratio.

#2) mol of NaOH used up by soy oil=mol of NaOH present initially - mol of remaining NaOH neutralized by HCl

=3.9 mol- (molarity of HCl*volume of HCl)=3.9 mol -(2.5mol/L*0.00920L)=3.877 mol NaOH or OH

Similrly,

mol of NaOH used up by butter=mol of NaOH present initially - mol of remaining NaOH neutralized by HCl

=3.9 mol- (molarity of HCl*volume of HCl)=3.9 mol -(2.5mol/L*0.00880L)=3.878 mol NaOH or OH

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