Question

In: Chemistry

15- Propane gas flows into a combustion chamber at a rate 250 L/min at 2.0 atmand...

15- Propane gas flows into a combustion chamber at a rate 250 L/min at 2.0 atmand ambient temperature. Air is added to the chamber at 1.0 atm and the same temperature. The gases are ignited.

a) To get complete combustion of the propane to carbon dioxide and water three times as much oxygen as is stoichiometrically appropriate is required. Assuming air is 21% oxygen and 79% nitrogen, calculate the required flow rate of air?

b )Under the conditions in part a, the combustion is not complete and a mixture of carbon dioxide and carbon monoxide is produced. It is determined that 94.0% of the carbon in the exhaust gas is present as carbon dioxide. The remainder is carbon monoxide. Calculate the percent composition of the exhaust gas in terms of mole fraction of CO, CO2 , O2 , N2 , and H2O. Assume the propane is completely reacted and the nitrogen is totally unreacted.

Can you please explain part B? I got 476.190 mL/min for part A.

Solutions

Expert Solution

a) Flow rate of propane gas is 250 L/min at 2.0 atm

assuming 1 min flow we have 250 L of propane

we have from PV=nRT

2 atm x 250 L = n x 0.082057 L atm K−1 mol−1 x 298 K

n = 20.447 moles

as per the stoichiometry of combustion propane we have the equation

C3H8 + 5O2 3CO2 + 4H2O

As per stoichiometry 1 mole propane requires 5 moles of oxygen it says in the problem that

To get complete combustion of the propane to carbon dioxide and water three times as much oxygen as is stoichiometrically appropriate is required.

so three time 5 is 15 times the number of moles of propane; so we have 15 x 20.447 = 306.711 moles of oxygen

from PV=nRT we get

1 atm x V = 306.711 x 0.082057 L atm K−1 mol−1 x 298 K

V = 7499.99 L This is assuming it is 100% but since air has only 21% oxygen we get

V = 35714.28 L/min

35714.28 L/min is the required air flow for complete combustion of the propane

B)

as per the stoichiometry of combustion propane we have the equation

C3H8 + 5O2 3CO2 + 4H2O

As per this stoichiometry we have 1 mole of propane gives 3 moles of carbon dioxide.

Now if only 94% of the carbon is present as carbon dioxide

6% less carbondioxide is formed

MW of carbon dioxide is 44 so we have

3 x 44 = 132 g of carbon dioxide /mole of propane

@ 94% it becomes 124 g of carbon dioxide /mole of propane

remaining 6% CO2 is present as carbon monoxide

((6% of 132 g)/44) x 28 = 5.08 g of carbon monoxide /mole of propane

Now the total propane is 20.447 moles

So carbon dioxide is 124 g x 20.447 = 2535 g or 57.61 moles

carbon monoxide is 5.08 x 20.447 = 103.87 g or 3.71 moles

water is 4 moles/mole of propane which is 4 x 20.447 = 81.79 moles

Air flow from part a is 35714.28 L/min which is 21% oxygen and 79% nitrogen; oxygen is 7499.99 L and nitrogen is 28214 L/min.

Moles of oxygen from part a is 306.711

amount of oxygen in 2535 g of carbon dioxide is 2535/44 x 32 = 1843 g

amount of oxygen in 103.87 g of carbon monoxide is 103.87/28 x 16 = 59.35 g

amount of oxygen in 80.79 moles of water is 1292 g

total oxygen present in carbon dioxide, carbon monoxide and water is 3194 g which is 99.83 moles

So the unreacted oxygen is 206.87 moles

28214 L of nitrogen is

PV = nRT

1 atm x 28214 L = n x 0.082057 L atm K−1 mol−1 x 298 K

n = 1153.8 moles which is 32306 g

percent composition of the exhaust gas in terms of mole fraction of CO, CO2 , O2 , N2 , and H2O.

moles of CO is 3.71 moles

moles of CO2 is 57.61 moles

moles of O2 is 206.87 moles

moles of N2 is 1153.8 moles

moles of H2O is 81.79 moles

total moles is 1503.78

mole percentage of

CO is 0.246 mole%

CO2 is 3.83 mole%

O2 is 13.75 mole%

N2 is 76.67 mole%

H2O is 5.43 mole%


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