Question

test ngine is fed octane (C8H18) fuel at a ae of 1.50 molis along with 120 mol of har .Both are fed at 12 atm pesure and 50C.Examinatioe of the exhaust gas reveals that 4.0% of octane did not react and thar CO is present with a CO/CO2 molar ratio of 1/5

Write two independent stoichiometric to account for the combustion of octant in this situation , reflecting the obsorved products

Answer 1

given

feed:1.5 moles of octane & 120moles of air

so moles of oxygen in feed=0.21*120=25.2moles

remaining 94.8 moles are of N₂

reaction stoichiometry

C8H18 +12.5O₂ -----> 8 CO₂+9H₂O                    .........(1)

C8H18+8.5O₂-------->8CO+9H₂O                         .............(2)

componant	feed stream	product stream
octane	1.5moles	0.06 moles
N2	94.8moles	94.8moles
O2	25.2moles	8.16moles
CO	0moles	1.92moles
CO2	0moles	9.6moles

• as 4% octane remain unreacted it will apear in product stream

so octane in product stream = 0.04*1.5 = 0.06 mol

• N2 does not take part in reaction so it will appear in product stream
• given that CO/CO₂=1:5

from stoichiometry 1 mol octane 8 mol CO8mol CO₂

so therefore amount of reacted octane reacts 5 times to give CO2 (reaction 1) that of reacted to produce CO by reaction 2

total amount of octane reacted =1.5-0.06=1.44

1.44/6=0.24

therefore 0.24 moles of octane undergoes reaction 2 to produce equivalent CO

&1.2 moles of octane react to give CO2 by reaction 1

therefore

0.24 moles of octane 8*0.24 CO1.92 moles of CO

1.2moles of octane8*1.2moles of CO₂9.6 moles of CO₂

by stoichiometry

O₂ reacted

by reaction 1

1.2 moles of actane12.5*1.2 moles of O₂15 moles of O₂

by reaction 2

0.24 moles of octane 8.5*0.24 moles of O₂2.04 moles of O₂

therefore total O₂ reacted = 15+2.04=17.04moles

unreacted O₂ will appear in product stream

unreacted O₂=25.2-17.04=8.16 mol