A 10.00mL sample of alcoholic ethyl acetate was diluted to 100.00 mL. 20.00 mL was aliquoted and mixed with 40.00 mL of 0.04672 M KOH. The resulting mixture was heated for 2 hours. CH3COOC2H5 + OH- → CH3COO- + C2H5OH After cooling, the excess OH was back titrated with 3.41 mL of 0.05042 M H2SO4.   Answer the following: Calculate the number of moles of OH- that reacted with ethyl acetate. Calculate the number of moles of ethyl acetate in the 20.00 mL solution. What is the mass of ethyl acetate (FW=88.11 g/mol) in the original 10.00 mL sample?

Fundamentals Of Analytical Chemistry
9th Edition
ISBN:9781285640686
Author:Skoog
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Chapter16: Applications Of Neutralization Titrations
Section: Chapter Questions
Problem 16.20QAP
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A 10.00mL sample of alcoholic ethyl acetate was diluted to 100.00 mL. 20.00 mL was aliquoted and mixed with 40.00 mL of 0.04672 M KOH. The resulting mixture was heated for 2 hours.

CH3COOC2H5 + OH- → CH3COO- + C2H5OH

After cooling, the excess OH was back titrated with 3.41 mL of 0.05042 M H2SO4.

 

Answer the following:

  1. Calculate the number of moles of OH- that reacted with ethyl acetate.
  2. Calculate the number of moles of ethyl acetate in the 20.00 mL solution.
  3. What is the mass of ethyl acetate (FW=88.11 g/mol) in the original 10.00 mL sample?
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