Propane is burned completely with excess oxygen. the product gas contains 24.8 mole% co2, 6.12% co, 40.8% h2o, and 28.6% o2.

a) calculate the percentage excess o2 fed to the furnace.

b) a student wrote the stiochemtric equation of the combustion of propane to form co2 and co as

2c3h8 + 17/2o2 > 3co2 + 3co +8h20

according to this equation, co2 and co should be a ratio of 1/1 in the reaction products, but in the product gas of part (a) they are in a ratio of 24.8/6.12. is the result possible? (hint: yes) explain how.

66% excess of O₂

Explanation:

a. The combustion of propane (C₃H₈) produce:

C₃H₈ + 5O₂ → 3CO₂ +4H₂O (1)

And:

C₃H₈ + ⁷/₂O₂ → 3CO + 4H₂O (2)

The sum of the two reactions is:

2C₃H₈ + ¹⁷/₂O₂ → 3CO₂ + 3CO + 4H₂O (3)

As 8 moles of H₂O are produced from 2moles of C₃H₈ the initial moles of propane (assuming 100 total initial moles) is:

40,8molH₂O× = 10,2 molesC₃H₈

For a complete reaction of 10,2 moles of C₃H₈ there are necessaries:

10,2 molesC₃H₈× = 43,35 mol O₂

As the product gas contains 28,6 moles of O₂, the initial total moles are:

43,35mol + 28,6mol = 71,95 initial moles of O₂

The formula of percentage excess O₂ is:

×100

Replacing:

×100 = 66% excess of O₂

b. The production of CO₂ and CO from C₃H₈ could be seen as two different reactions (1) and (2). The different conditions of combustion of propane could produce more CO₂ over CO and vice versa. For this reason, the ratio of CO₂/CO could be different of 1.

I hope it helps!

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of c is as shown here in the image.

c (diamond),

sio2 (quartz)