Question

prin reactia a 280 m3 etena ( masurat in conditii normale ) de puritate 80% cu h20 se obt etanol. acesta se dizolva in apa rezultand 1699 kg de c=23%. Calculati randamentul de transformare a etenei in etanol.

Answer 1

C2H4 + H2O = C2H5OH
volum pur = 280*80/100 = 224 m3
22,4 l etena..........46 g etanol
224 m3 etana.............x kg
x = 460 kg 
c = md/ms * 10 => md = 1699*23/100=390.77 kg
randamentul = masa practica/masa teoretica * 100 
= 390.77/460 * 100 = 84.95%

Answer 2

n=V/Vm=280/22.4=12.5 kmoli CH2=CH2

0.8*12.5=10 kmoli pur C2H4

n=m/M==>m=n*M=10*28=280 kg pur etena


28kg...................................46kg
CH2=CH2 + H-OH ---> CH3-CH2(OH)
280kg................................x=460kg

Cp=md/ms*100==>md=Cp*ms/100=1699*23/100=390.77g C2H6O

n(eta)=mp/mt*100=390.77/460*100=84.95%

MC2H4=28kg/kmol
MC2H6O=46kg/kmol

Vm=Volumul molar = 22.4m3 (c.n)