Question

0 Use of the Data Booklet is relevant to this question.

Ethyl ethanoate can be obtained from ethanoic acid and ethanol by the following reaction.

CH3CH2OH + CH3CO2H CH3CO2CH2CH3 + H2O

Ethanol (30 g) and ethanoic acid (30 g) are heated under reflux together, and 22 g of ethyl
ethanoate are obtained.

What is the yield of the ester?
A 25 % B 38 % C 50 % D 77 %

Answer 1

this is just stoichiometry. Convert the reactants to moles - determine what the limiting reactant (LR) is - use the moles of the LR to find the theoretical moles produced with the ratio. Convert to grams and compare to the actual amount you obtained (given in the question).

Answer 2

i did that 30/46=0.65 moles of ethanol
30/60=0.50 moles of ethanoic acid
22/88=0.25 moles of ethyl ethanoate

Answer 3

the % yield of the ester is in grams (usually) so dont convert that to moles.

Answer 4

22*2/1=44 grams
22/44=0.5*100=50%

Answer 5

thank you :)

Answer 6

no problem. So 44 grams is your theoretical yield for the ester?

Answer 7

nevermind, it's good