10 Calculate the amount of ester, CH;COOC,Hs formed when 2.00 moles of ethanoic acid, CH3COOH and 2.00 moles of ethanol, C2H5OH are allowed to come to equilibrium at 300 K. CH3COOH(aq) + C2H$OH(aq) = CH3COOC2H5(aq)+ H2O(1) Kc = 4.12 (ANS = NCH3COOC2H5= 1.414 mole; mass CH3COOC2H5= 124.43 g) N4. HS c NHR Cg
10 Calculate the amount of ester, CH;COOC,Hs formed when 2.00 moles of ethanoic acid, CH3COOH and 2.00 moles of ethanol, C2H5OH are allowed to come to equilibrium at 300 K. CH3COOH(aq) + C2H$OH(aq) = CH3COOC2H5(aq)+ H2O(1) Kc = 4.12 (ANS = NCH3COOC2H5= 1.414 mole; mass CH3COOC2H5= 124.43 g) N4. HS c NHR Cg
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![(ANS = [H2] & [I2] = 0.700mol dm3; [HI] = 0.0992 mol dm³)
10 Calculate the amount of ester, CH3COOC2H; formed when 2.00 moles of ethanoic
acid, CH3COOH and 2.00 moles of ethanol, C2H5OH are allowed to come to
equilibrium at 300 K.
CH;COOH(aq) + C2H$OH(aq)
- CH3COOC2H5(aq)+ H2O(1)
Kc = 4.12
(ANS = 1CH3COOC:Hs= 1.414 mole; mass CH3COOC:H5= 124.43 g)
%3D
NH, HS esl NH g
nge
n.x
*-0.33](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58e0763f-97b5-4605-a042-68bac79de385%2F9f12cd03-3d89-43d6-a94c-c6eb5d1cae30%2Fwuxr5qr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(ANS = [H2] & [I2] = 0.700mol dm3; [HI] = 0.0992 mol dm³)
10 Calculate the amount of ester, CH3COOC2H; formed when 2.00 moles of ethanoic
acid, CH3COOH and 2.00 moles of ethanol, C2H5OH are allowed to come to
equilibrium at 300 K.
CH;COOH(aq) + C2H$OH(aq)
- CH3COOC2H5(aq)+ H2O(1)
Kc = 4.12
(ANS = 1CH3COOC:Hs= 1.414 mole; mass CH3COOC:H5= 124.43 g)
%3D
NH, HS esl NH g
nge
n.x
*-0.33
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