Question

100 moles/hour of pure ethane at 25 degrees Celsius and 1.0 atm are burned with 50.0% excess air producing a desired product of carbon dioxide and an undesired product of carbon monoxide. The fractional conversion of the ethane is 80.0% (on a molar basis). 95% of ethane that reacts goes to carbon dioxide. Determine the molar flow rate and molar composition of the gaseous product stream, which is at 400 degrees Celsius and 1.0 atm.

Completely label your diagram, list the temperature, pressure, flow rate, and composition (mole fractions with indexes) for eac stream, supply values and units for knowns and question marks for unknowns.

Chemical Reactions:
I: __C2H6 + __O2 ---> __CO2 + __ H2O
II: __C2H6 + __O2 ---> ___CO + ___H2O

Answer 1

Basis : 100 moles/hr of pure ethane. Combustion of ethane I s represented as

C2H6+ 7/2O2-->2CO2 +3H2O

1 moles of ethane requires 3.5 moles of O2

100 moles/hr requires 100*3.5= 350 moles/hr of oxygen

Air contains 21%O2 and 79%N2, moles of air =350/0.21=1667 moles/hr

Air supplied is 50% excess, moles of air supplied =1.5*1667=2500.5 moles/hr

Total moles of Feed = 100+525(O2)+1975.5 (N2) = 2600.5 moles/hr

Feed composition : C2H6 = 100/2600.5= 0.0384, O2= 525/2600.5=0.202, N2= 1975.5/2600.5=0.76

Oxygen supplied =2500.5*0.21=525 moles/hr, N2= 2500.5-525= 1975.5 moles/hr

Fractional conversion of ethane is 80%. So moles of ethane converted = 80 moles

Out of this 80 moles, 95% reacted with oxygen to give CO2.

Moles of Ethane reacted to form CO2= 80*0.95=76 moles/hr

Moles of CO2 formed = 2*76=152 moles/hr

Moles of H2O formed =3*76= 218 moles/hr

Oxygen consumed = 76*3.5 =256 moles/hr

Rest of 80-76 moles gives   CO after reaction with oxygen

C2HJ4+2O2->2CO +2H2O

Moles of CO formed = 2*4= 8 moles/hr, moles of O2 consumed = 2*2= 4 moles

Moles of H2O formed =2*2= 4 moles/hr

Products Contains ( moles/hr)

C2H6= 100- 80=20 moles/hr

CO2= 152 moles/hr

CO= 8 moles/hr,

H2O= 218+4=222 moles/hr

O2= Oxygen supplied- (oxygen consumed for CO2 formation + Oxygen supplied for CO formation)

= 525- ( 256+4)= 265

N2=1975.5

Total molar flow rate of products = 20+152+8+222+265+1975.5 = 2542.5 moles/hr

Composition : C2H6= 20/2542.5 = 0.007866, CO3= 152/2542.5 = 0.0575, CO= 8/2542.5 = 0.003147, H2O= 222/2542.5=0.0873, O2: 265/2542.5=0.1042, N2= 1975.5/2542.5=0.7475