100 mol/h of propane (C_3H_8) is burned with 35% excess air in a furnace. Only a
ID: 1843039 • Letter: 1
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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