\"Propane is burned completely with excess oxygen. The product gas contains 24.8

Question

"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 stoichiometric equation of the combustion of propane to form CO2 and
CO as
2C3H8 + 17/2 O2 ---> 3CO2 + 3CO + 8H2O
According to this equation, CO2 and CO should be in 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 that result possible? (Hint: Yes.) Explain how."

Explanation / Answer

2C3H8 + (17/2)O2   ---------> 3CO2 + 3CO + 8H2O

Let the total moles of the gas mixture be 100

Thus, moles of propane = 0

moles of oxygen = 28.6

moles CO = 6.12

moles of CO2 = 24.8

Moles of water = 40.8

Now, moles of CO2 that must form according to balanced reaction = (3/8)*moles of H2O formed = 15.3 = moles of CO formed as per balanced reaction

Thus, moles of excess CO2 produced = 24.8 - 15.3 = 9.5

Hence % excess of O2 = {(molesof CO2/2)/100)*100 = 4.75 %

b) yes, the ratio 24.8:6.12 is possible because if excess of O2 is present the CO produced will be further oxidized to CO2 according to the equation:-

2CO + O2 ----------> 2CO2

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