Butane (C4H10) is to be burned with air. If 10 mol butane per second are fed wit

Question

Butane (C4H10) is to be burned with air.

If 10 mol butane per second are fed with 45% excess air, determine the molar flowrate and composition of the product gases under the following conditions:

     - 98% conversion of butane

     - No carbon monoxide produced

Solve using mole balances

Butane (C_4H_10) is to be burned with air. If 10 mol butane per second are fed with 45% excess air, determine the molar flowrate and composition of the product gases under the following conditions: 98% conversion of butane No carbon monoxide produced Solve using mole balances

Explanation / Answer

10 mol of butane needs 65 moles of O2

65 mol O2 is present in (65 mol / 0.21 ) mol of air = 309.5 mol air

45 % excess of air

total air fed = 309.5 + 0.45 x 309.5 = 448.8 mol

98% conversion.

So moles of butane reacted = 9.8 moles

Moles of butane left =0.2 moles

Moles of O2 reacted = 0.98 x 65 = 63.7 moles

moles of O2 left == ( 0.21 x 448.8 mol - 63.7 moles) ==30.548 moles

moles of CO2 formed = 4 x 9.8 mol = 39.2 mol

moles of water formed = 5 x 9.8 = 49 moles

nitrogen in air doesnot participate in reaction

moles of N2 = 0.79 x 448.8 mol = 354.55 moles

for molar flow rates keep per second for calculated moles

Final composition is

butane = 0.2 mol

nitrogen = 354.55 mol

O2 = 30.548 moles

CO2 = 39.2 mol

water = 49 moles

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