Question

Volumul de etena masurat in conditii normale de presiune si temperatura , necesar obtinerii a 1240 g etandiol este :

Answer 1

28g...................................62g
CH2=CH2 + 2H2O --> CH2(OH)-CH2(OH) + H2
x=560g.............................1240g

n=m/M=560/28=20 moli C2H4

n=V/Vm==>V=n*Vm=20*22.4=448L C2H4

MC2H4=28g/mol
MC2H6O2=62g/mol