Question

Chemical engineering question

A synthesis gas containing 6.4% CO₂, 0.2% O₂, 40.0% CO and 50.8% H₂ (the remainder is N₂) is burned with 40 % excess dry air (assume 21 % O₂ and 79 % N₂).

i)Identify the reactions occurring during combustion of this syngas.

ii)Calculate the amount of air added per 100 mol of syngas.

iii)If complete combustion occurs, then calculate the composition of the flue gas on a weight basis.

Answer 1

A synthesis gas containing 6.4% CO₂, 0.2% O₂, 40.0% CO and 50.8% H₂ (the remainder is N₂) is burned with 40 % excess dry air (assume 21 % O₂ and 79 % N₂)

Solution

Composition of synthesis gas =

The moles of syn gas present = 100moles

So percentage of gases in the syn gas will be

CO2 = 6.4% = 6.4moles

O2 = 0.2% = 0.2 moles

CO = 40 % = 40 moles

H2 = 50.8 % = 50.8 moles

N2 = 100- (50.8 + 40 + 0.2 + 6.4 ) = 2.6% = 2.6 moles

H2 and CO will undergo combustion, the reactions will be

CO + 1/2O2 ---> CO2

H2 + 1/2O2 ---> H2O

So moles of oxygen required for CO= 0.5 X 40 = 20 moles

moles of oxygen required for H2 = 0.5 X 50.8 = 25.4 moles

Total moles of oxygen required = 20 + 25. 4 = 45 .4 moles

Moles already present in syn gas = 0.2 moles

so actual oxygen required = 45.4 - 0.2 = 45.2 moles

each mole of air contains 0.21 moles of O2

So moles of air required for 45.2 moles of O2 = 45.2 / 0.21 = 215.238 moles

The air used is 40% excess

So air supplied = 140 X 215.238 / 100 = 301.333 moles

Moles of O2 supplied =301.333 moles X 0.21 = 63.2799 moles

flue gas composition

The flue gas composition = Moles of N2 + Moles of O2 + Moles of CO2 + moles of H2O

N2 left after combustion = N2 present initially in the sample + N2 from air

                                       = 301.3332 X 0.79 + 2.6 = 240.653 moles

O2 left after combusion = 63.2799 - 45. 4 = 17.8799 moles

moles of CO2 formed = Mole sof CO2 present + moles of CO2 formed = 6.4 +40 = 46.4 moles

moles of H2O formed = 50.8 moles

total moles = 240.653 moles + 17.8799 + 46.4 + 50.8 = =355.733 moles

let us calcualte the percentage of each gas

O2 = Moles of O2 X 100 / total moles = 17.8799 X 100 / 355.733 = 5.03%

N2 = Moles of N2 X 100 / total moles = 240.653 X 100 / 355.733 = 67.65%

H2O = Moles of H2O X 100 / total moles = 50.8 X 100 / 355.733 =14.28 %

CO2 = Moles of CO2 X 100 / total moles = 46.4 X 100 / 355.733 =13.04%