Answer to Question #144792 in General Chemistry for Joshua

Question

Butane (C4H10) is burned with 35% excess air. Find the composition of the flue gas if combustion is complete.

Answer 1

2C4H10 + 13O2 = 8CO2 + 10H2O

for burned 1 mol of butane, we need 6.5 mol of O2. If we have 35% excess than 1 mol of butane we waste 8.775 mol of O2. Because air it's not clean O2, but 78% N2, 21% O2, and 1% Ar, so we waste 41.786 mol of air for 1 mol butane, or 8.775 mol O2, 32.593 mol N2, and 0.418 mol Ar. So in flue gas(product of 1 mole C4H10) we will have 4 mol CO2(44 g/mol), 5 mole H2O(18 g/mol), 2.275 mole O2(32 g/mol), 32.593 mole N2(28 g/mol), and 0.418 mole Ar(39.95 g/mol). Or in percent 13.88% CO2, 7.1% H2O, 5,74% O2, 71.97% N2, 1.31% Ar.