C6H5-CH2-OH + CH3-COOH ⇒C6H5-CH2-OCOCH3 + H2O
M ester = 150 g/mol⇒
n ester = n alcool reactionat = n acid reactionat = 1mol
80% ·n initial alcool = 1mol ⇒ n initial alcool = 1,25 moli (a)
n initial acid = 2 + 1 = 3 moli (d)
K = [ester]·[apa] /{[alcool]·[acid]} la echilibru
K = 1·1 / 0,25·2 = 2 (c)
C acid = 1/3 ·100 = 33,3 % (e)