A mixture of 72 mole% butane with remainder hydrogen undergo combustion with 25%

Question

A mixture of 72 mole% butane with remainder hydrogen undergo combustion with 25% excess air. The percent conversion is 85% for both hydrogen and butane. Of the butane that reacts, 90% reacts to form carbon dioxide with the remainder reacting to form carbon monoxide. Assume a basis of 5 moles of the fuel mixture. How many moles of air are fed to the process?

Explanation / Answer

C4H10 + 6.5 O2 --->4CO2 +5 H2O C4H10 + 4.5O2 ----> 4CO +5 H2O H2 + 0.5 O2 ---> H2O. moles of O2 = 5(0.72 x 6.5 x 0.9 x100/85) +5(0.72 x 4.5 x0.1x100/85) + 5(0.28 x 0.5x100/85) = 24.776+1.9+0.8235 = 27.5 , air has 21 % Oxygen , so air required = (27.5/21) x 100 = 130.95 moles

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