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

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Chapter1: Chemical Foundations
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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$OH(aq)
- CH3COOC2H5(aq)+ H2O(1)
Kc = 4.12
(ANS = 1CH3COOC:Hs= 1.414 mole; mass CH3COOC:H5= 124.43 g)
%3D
NH, HS esl NH g
nge
n.x
*-0.33
Transcribed Image Text:(ANS = [H2] & [I2] = 0.700mol dm3; [HI] = 0.0992 mol dm³) 10 Calculate the amount of ester, CH3COOC2H; formed when 2.00 moles of ethanoic acid, CH3COOH and 2.00 moles of ethanol, C2H5OH are allowed to come to equilibrium at 300 K. CH;COOH(aq) + C2H$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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