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!

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.