I need help with getting the calculations right, I am not just asking for the an

Question

I need help with getting the calculations right, I am not just asking for the answers, I want to understand it. I have a test coming up and this will be in it, so I need to know how to get each answer.

Thank you.

1) Volume of 0.0100 M NaOH used in the WVO titration ________5.9 ml___________

2) Moles of NaOH needed to neutralize the free fatty acids in 1.00 mL of

Waste Vegetable Oil (WVO)__5.9*10-5 mol___________      

3) Moles of KOH you will need to neutralize the free fatty acids in 10.0 mL of

WVO_____________________

The amount of base needed as a catalyst for the transesterification reaction is equal

to 3.5 grams of NaOH per 1000 mL of PURE veggie oil.

4) How many grams and moles of NaOH will be needed as catalyst for 10.0 mL of

PURE veggie oil?

_____________________________grams ____________________________moles

5) We are using KOH as catalyst. How many moles of KOH will be needed for 10.0

mL of PURE veggie oil?

___________________________moles KOH catalyst for 10.0 mL pure veggie oil

6) NOW, how many TOTAL moles of KOH will be needed for the transesterification

reaction: moles KOH to neutralize fatty acids + moles KOH as reaction catalyst =

___________________                    total moles

7) The KOH for transesterification is provided as a 10.0 M aqueous solution. How

many mL, micro liters and deci liters of 10.0 M KOH will be needed?

______________________________mL _____________________________ mL____________________dL

8) Should you have used refined vegetable oil instead of WVO, how much 10 M KOH you

       would have used?

Explanation / Answer

Titration

1. WVO containing free fatty acids. Each 1 mole of free fatty acid require 1 mole of NaOH for neutralization,

R-COOH (free fatty acid) + NaOH --> R-COONa + H2O

2. molarity = moles/L

So, moles = molarity of NaOH x volume of NaOH used (L)

3. so when 1 ml WVO required x moles (found in 2. above) of NaOH,

10 ml of WVO would need = 10 x x moles of KOH

If molarity of NaOH = molarity of KOH

Volume of NaOH used should be = volume of KOH needed for the reaction.

4. Given 3.5 g NaOH is needed as catalyst for 1000 ml of vegetable oil transesterification

So, 10 ml vegetable oil would need = 3.5/100 = 0.035 g of NaOH

moles of NaOH = 0.035 g/40 g/mol = 0.000875 mol

5. moles of KOH needed for transesterifcation,

= 0.035 g/56 g/mol = 0.000625 mol

6. Total moles of KOH needed = 5.9 x 10^-5 + 6.25 x 10^-4 = 6.84 x 10^-4 mols

7. ml of KOH needed = 6.84 x 10^-4 mol/10 M = 6.84 x 10^-5 L

= 6.84 x 10^-2 ml

= 78.4 uL

= 6.84 x 10^-4 deciliter

8) Refined vegetable oil has less free fatty acid and would requires less amount of KOH for the reaction.                 

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.