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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 (where the = is the double arrow arrow)
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 is C, I need the explanation or workout for the calculations please!
Convert the masses of ethanol and ethanoic acid to moles to figure out the limiting reagent.
Limiting reactant is acid, 0.5 moles. Expected moles of ester is thus 0.5 moles, 44g.
%yield = 22/44 = 50%