CH3COOH +C2H5OH⇄ CH3COOC2H5 +H2O
mol initial 1. 0,4 0 0
mol transformati -x -x. +x +x
mol la ech. 1-x 0,4-x x x
K= [ester}[apa]/[acid][alcool]= x² / (1-x)(0,4-x)
x= 0,35mol
conversie acid in ester= 0,35x100/1= 35%
conversie alcool in ester= 0,35x100/0,4=87,5%
verifica, te rog, rezolvarea ecuatiei de ordinul II !!!!!