vom scrie reacția:
CH₃-COOH + CH₃-CH₂-OH ⇔ CH₃-COOCH₂CH₃ + H₂O
Macid=MC₂H₄O₂=12×2+4+16×2=24+4+32=60g/mol
Malcool=MC₂H₆O=12×2+6+16=24+6+16=46g/mol
Mester=MC₄H₈O₂=12×4+10+16×2=48+8+32=88g/mol
n=m/M=176/88=2moli ester
Stim ca la echilibru aceeasi cantitate de reactanti se transforma in produsi de reactie. => se transforma 2 moli de alcool si 2 moli de acid
Acidul are Ct=50% => 50/100=ntrans/ntotal => => ntotal=ntrans×100/50=2×ntrans=2×2=4 moli acid
ptr alcool:
2 moli........100%
x moli........150% (cu tot cu exces)
x=2×150/100=3moli alcool
m acid=4×60=240g
m alcool=3×46=138g
[tex] \frac{92g}{100ml} = \frac{138g}{ V_{alcool} }[/tex] => Valcool=138×100/92=150ml
[tex] \frac{92,3g}{100} = \frac{240g}{ V_{acid}}[/tex] => Vacid=240×100/92,3=260ml
