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In a furnace, 76 lbs propane C3H8 are burned with 17.2% excess air for complete combustion....

In a furnace, 76 lbs propane C3H8 are burned with 17.2% excess air for complete combustion. What is the percent CO2 in combustion products? Round your answer to the one decimal place.

Expert Solution

Complete combustion of propane

C3H8 + 5(O2 + 3.76N2) = 3CO2 + 4 H2O + 18.8 N2

Moles of propane = mass/molecular weight

= 76lb / 44.1lb/lbmol

= 1.723 lbmol

Theoretical moles of O2 required = 1.723 x 5 = 8.617 lbmoles

Theoretical air required = 8.617/0.21 = 41.033 lbmol

Actual air supplied = 41.033 + 41.033*17.2/100 = 48.1 lbmol

Moles of O2 in air = 0.21 x 48.1 = 10.101 lbmol

Moles of N2 in air = 37.9 lbmol

In the product

Moles of CO2 formed = 1.723*3 = 5.169 lbmol

Mass of CO2 = moles x molecular weight

= 5.169 lbmol x 44 lb/lbmol = 227.436 lb

Moles of H2O = 1.723*4 = 6.892 lbmol

Mass of H2O = 6.892 lbmol x 18 lb/lbmol = 124.056 lb

Moles of N2 = 37.9 lbmol

Mass of N2 = 37.9 lbmol x 28 lb/lbmol = 1061.2 lb

Moles of O2 = actual - theoretical = 10.101 - 8.617

= 1.484 lbmol

Mass of O2 = 1.484 lbmol x 32 lb/lbmol = 47.488 lb

Total mass of products

= 227.436 + 124.056 + 1061.2 + 47.488

= 1460.18 lb

Mass % of CO2 = mass of CO2 x 100 / total mass

= 227.436*100/1460.18

= 15.6 %

Amanda K answered 2 years ago

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