Question

Se supun esterificarii 1, 781 moli acid acetic si 3,5 moli etanol.
1. Denumeste esterul obtinut si calculeaza compozitia in procente de masa a amestecului supus esterificarii.
2. Calcuzeaza constanta de exhilibru a reactiei de esterificare daca la echilibru se gasesc 1.5 moli ester.
Multumesc♥

Answer 1

CH3-COOH + HO-CH2-CH3 = CH3-CO-O-CH2-CH3 + H2O
Acetat de etil
M acid acetic = 60 g/mol
M etanol = 46 g/mol
n = m/M
m etanol = 46*3,5 = 161 g etanol
m acid = 1,781*60 = 103,08 g acid acetic
I    1,781 moli        3,5 moli                  -             -
C      x                       x                          -           -
E   1,781-x             3,5-x                       x            x
x = 1,5 moli ester format
n etanol = 3,5-x = 2 moli
n acid = 1,781-1,5 = 0,281 moli
K = ([H2O]*[ester])/([acid]*[alcool]) = (1,5/V*1,5/V)/(2/V*0,281/V) = 4
K = 4