-calculez masa etanol din solutie
c= mdx100/ms, unde ms= roxVs
90/100= md/ 0,85x1400------>md=1071g
-calculez din ecuatie ,mol etena obtinuti teoretic
1mol ....................1mol
CH3-CH2OH----> CH2=CH2+H2O
46g.......................1mol
1071g.....................n=23,28mol......teoretic
dar n,practic= 20mol,mai putin din cauza pierderilor
randamentul = 20x100/23,28= 85,9%