100 mol/h of propane (C_3H_8) is burned with 35% excess air in a furnace. Only a

Question

100 mol/h of propane (C_3H_8) is burned with 35% excess air in a furnace. Only a 92% conversion of the propane is achieved. Of the propane that reacts, 85% reacts to form CO_2 while the remaining forms CO (incomplete combustion). Write balanced equations for both reactions. Determine the rate at which O_2 and N_2 supplied to the furnace (mol/h). remember that %XS is calculated based on assuming complete conversion of the limiting reactant, for combustion reactions, this is based on complete combustion. What is the mole % (on a dry basis, omit water) of CO_2 leaving the reactor?

Explanation / Answer

Solution:

C3H8+5O2----> 3CO2+4H2O and

C3H8+3.5O2----> 3CO+ 4H2O

Basis :

100mol/hr propane requies 5 times the oxygen for complete combustion = 5 times =5×100 mol/hr= 500 mol/hr

Moles of air to be supplied =500/0.21 (oxygen is 21% of air)= 2381 moles/hr for complete combustion

Air contains

79% N2 and

21%, Oxygen :

O2= 500mole/hr and

N2= 2381-500 =1881moles/hr

Air is suppliled 35% excess,

Oxygen supplied = 500×1.35 =675 moles/hr N2=1881×1.35=2539.35 moles/hr

Out of 100 moles/hr 85 moles react with oxygen to from CO2. Moles of CO2 formed= 3×85=255 moles/hr

Oxygen consumed for formation of CO2= 85×5=425 moles/ hr(1)

Remaining 15 moles/hr of propane gives CO, amount of CO formed = 3×15 =45 moles/hr

Oxygen cosnumed for this incomplete combustion =15×3.5= 52.5 moles/hr (2)

Oxygen remaining= Oxygen supplied- ( oxygen used for formation of CO2 (1)+Oxygen use for CO formation (2))

=675-(425+70)= 180 moles/hr

Products ( on dry basis ): N2 = 2539.35moles/hr

O2( unreacted + excess)= 180 moles/hr

CO2=255 moles/hr

CO= 45 moles/hr

Total moles of product= 2539.35+180+255+45 = 3019.35 moles/hr

CO2 % in the gases leaving = 100×255/3019.35=8.44%

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