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A synthesis gas analyzing 6.4% CO2, 0.2% O2, 40.0% CO, and 50.8% H2, (the balance is...

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?

Expert Solution

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%

queen_honey_blossom answered 2 years ago

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