Question

30lb of coal (83%Carbon, 1%Sulfur, 12%Hyrdogen and 4%H20) are burned with 600lb of air yielding a gas in which ratio of CO2 to CO is 3 to 2. What is the mole percent of each component in the flue gas

Answer 1

Given the mass of air = 600 lb = 600 lb x (453.592 g / lb) = 272155 g

moles of air = 272155 g / 28.97 g/mol = 9394.4 mol

Moles of O2 in air =  9394.4 mol x 21/100 = 1972.8 mol O2

moles of N2 in air = 7421.6 mol N2

Given the mass of coal = 30 lb = 30 lb x (453.592 g / lb) = 13608 g

Mass of carbon in the coal = 13608 g x 83/100 = 11295 g

moles of carbon in the coal = 11295 g / 12.0 g/mol = 941.22 mol

moles of CO2 produced = 941.22 mol x 3/5 = 564.7 mol CO2

moles of CO produced = 941.22 mol x 2/5 = 376.5 mol CO

Mass of S in the coal = 13608 g x 1/100 = 136.08 g

moles of S in the coal = 136.08 g / 32.06 g/mol = 4.244 mol

Mass of H2 in the coal = 13608 g x 12/100 = 1633 g

moles of H2 in the coal = 1633 g / 2.0 g/mol = 816.5 mol

Mass of H2O in the coal = 13608 g x 4/100 = 544.3 g

moles of H2O in the coal = 544.3 g / 18.0 g/mol = 30.24 mol

  The chemical reaction for the combustion of coal is

C(s) + S(s) + H2 + H2O + O2 -------- > CO2(g) + CO(g) + SO2(g) + H2(g) + H2O(g)

Hence total moles of O2 required for combustion

= 564.7 mol (due to CO2) + 188.25 mol (due to CO) + 4.244 mol (due to S)

= 757.2 mol O2

Hence moles of O2 remain unreacted = 1972.8 mol - 757.2 mol = 1215.6 mol O2

Hence composition of flue gas:

=========================

7421.6 mol N2 + 1215.6 mol O2 + 564.7 mol CO2 + 376.5 mol CO + 4.244 mol SO2 + 816.5 mol H2 + 30.24 mol   H2O

Hence total moles of all gases in flue gas = 10429.4 mol

Hence

mole % of N2 = (7421.6 mol N2 / 10429.4 mol ) x 100 = 71.2 % (answer)

mole % of O2 = (1215.6 mol O2 / 10429.4 mol ) x 100 = 11.66 % (answer)

mole % of CO2 = (564.7 mol CO2 / 10429.4 mol ) x 100 = 5.41 % (answer)

mole % of CO = (376.5 mol CO / 10429.4 mol ) x 100 = 3.61 % (answer)

mole % of SO2 = (4.244 mol SO2 / 10429.4 mol ) x 100 = 0.041 % (answer)

mole % of H2 = (816.5 mol H2 / 10429.4 mol ) x 100 = 7.83 % (answer)

mole % of H2O = (30.24 mol H2O / 10429.4 mol ) x 100 = 0.290 % (answer)