Question

Se dau urmatorii acizi: A(acid metanoic) si B (acid etanoic). Acestia se afla in amestec in raport molar 1:1. Stiind ca amestecul se esterifica cu 1,78 moli de alcool etilic sa se afle masa amestecului de acizi si compozitia procentuala a amestecului.

Answer 1

Fiind raport de 1:1 => luam x:x
1mol...............1mol
H-COOH + CH3-CH2-OH <=> HCOOCH2CH3 + H20
x......................x
1mol.................1mol
CH3-COOH + CH3-CH2-OH <=> CH3COOCH2CH3 + H2O
x........................x
2x=1,78 => x=0,89moli
reamintesc: n=m/M
Mac.metanoic=12+2+16*2=14+32=46g/mol
Mac.etanoic=12*2+4+16*2=60g/mo
m=n*M
m=0,89*46=40,94g ac. metanoic
m=0,89*60=53,4g ac. etanoic
mamestec=40,94+53,4=94,34g amestec
94,34g.....................100%
40,94g.........................y%
53,4g...........................z%
y=40,94*100/94,34=43,4% metanoic
z=53,4*100/94,34=56,6% etanoic