Question

A synthesis gas analyzing 6.4% CO2, 0.2% O2, 40.0% CO, and 50.8% H2, (the balance is N2), is burned with 40% dry excess air. What is the composition of the flue gas?

Answer 1

The balanced equation will be based on the fact only hydrogen and CO will undergo combustion other gases will not

Let we have 100moles of synthesis gas

Moles of CO2 = 6.4

Moles of O2 = 0.2

Moles of CO = 40

Moles of H2 = 50.8

Moles of N2 = 100-(6.4+0.2+40+50.8) = 2.6

Combustion equation will be

H2 + 0.5O2 --> H2O [so moles of H2O formed =50.8]

CO + 0.5O2 --> CO2   [ so moles of CO2 formed = moles of CO = 40]

So moles of Oxygen required = 0.5 X 50.8 + 0.5 X 40 = 45.4 moles

Moles already present =0.2

Moles of oxygen required =45.2 moles

Excess oxygen required = 40 X 45.2 / 100 = 18.08 moles

So total moles of oxygen required = 45.2 + 18.08 = 63.28 moles

Moles of nitrogen alongwith Oxygen = 63.28 X 79 / 21 = 238.05

a) 6.4 moles of CO2 + 40 moles of CO2 (from CO) = 46.4 moles of CO2

b) moles of H2O from H2 = 50.8 moles

c) moles of N2 = 2.6 (already present) + 238.05 = 240.65

d) excess O2 = 18.08

Total moles of gases = 18.08 + 240.65 + 50.8 + 46.4 = 355.93 (100%)

So percentage composition of flue gas

a) CO2 = 46.4 X 100 / 355.93 = 13.04%

b) H2O = 50.8 X 100 / 355.93 = 14.27%

c) O2 = 18.08 X 100 / 355.93 = 5.08 %

d) N2 = 240.65 X 100 / 355.93 = 67.61%