d=ms/Vs => ms=d*Vs=0,8*225=180g solutie etanol
md=c*ms/100=171,9g etanol
n=m/M=171,9/46=3,74moli etanol
m=d*V=210g acid acetic
n=m/M=3,5moli
esterul este acetatul de etil
nester=m/M=260/88=2,9545moli
deoarece nacid<nalcool => avem exces de alcool deci vom folosi acidul
1mol............................................1mol
CH3-COOH + CH3-CH2-OH <=> CH3-COOCH2CH3 + H2O
3,5moli.........................................3,5moli
randament=n practic*100/n teoretic=2,9545*100/3,5=84,41%