CH3-COOH + CH3-CH2-OH ⇄ CH3-COO-CH2-CH3 + H2O
I: 3 2 1 1
C: x x x x
Ec: (3-x) (2-x) x+1 x+1
Kc=[acetat]*[H2O]/[etanol]*[acid acetic] ; 4=(1+x)²/(3-x)(2-x) ⇔ 4(3-x)(2-x)=(1+x)² ⇔ 4(6-5x+x²)=x²+2x+1 ⇔ 3x²-22x+23=0 ==> x1=1.26
3-1.26=1.74 moli CH3-COOH ; 0.74 moli CH3-CH2-OH ; 2.26 moli eter ; 2.26 moli H2O
nr moli amestec = 7
%acid acetic=1.74/7*100==>Calculeaza!!!
%etanol=0.74/7*100==>
%eter=2.26/7*100==>
%apa=2.26/7*100==>