Question

Continuare problema 3.10

Answer 1

V=100L
VCH4=25%=25m³
VC2H6=10%=10L
VC3H8=10%=10L
VC5H12=10%=10L (nereactionat)


22.4m³.....................................22.4m³..22.4m³
CH3-CH2-CH2-CH2-CH3 ---> C4H8 + CH4
25m³.........................................25m³.....25m³

22.4m³.........................................22.4m³............22.4m³
CH3-CH2-CH2-CH2-CH3 ---> CH3-CH=CH2 + CH3-CH3
10m³...........................................10m³...................10m³

22.4m³.........................................22.4m³............22.4m³
CH3-CH2-CH2-CH2-CH3 ---> CH2=CH2 + CH3-CH2-CH3
10m³...........................................10m³...................10m³

22.4m³..................................................22.4m³
CH3-CH2-CH2-CH2-CH3 ---> CH3-CH2-CH2-CH2-CH3
10m³....................................................10m³

In total 25+10+10+10=55m³ C5H12 ==> 100-55=45m³ reactionati

%n(eta)=Vr/Vt*100=45/55*100=81.81%

x%din 55m³(%de C5H12 transformat in CH4)=25m³

55x/100=25==>55x=2500==>x=45.45%

MC5H12=72kg/kmol

n=m/M=144/72=2 kmoli C5H12

55m³.................100m³ amestec
2kmoli.................x=3.636 kmoli

pV=nRT==>V=nRT/p=3.636*0.082*400/2=59.63636m³ amestec *1000=59636.36 L amestec