Question

A heater burns n-butanol with 60% excess air. Combustion is complete, the fuel is exhausted and the stack gas leaves at a temperature of 260 °C and a pressure of 100 kPa. What would be the results of an Orsat analysis of the flue gas? (Calculate both dry and wet basis.) How many liters of flue gas are produced per mole of n-butanol burnt?

Answer 1

The balanced equation for combustion of n- butanol is give by

C₄H₁₀O + 6 O2 --> 4 CO2 + 5 H2O

Basis : 1 mole of n-butanol oxygen required

Moles of oxygen required = 6 moles. Air contains 79% Nitrogen and 21% Oxygen, hence moles of air required= 6/0.21=28.57

Mole of air supplied is 60% excess, hence moles of air supplied= 1.6*28.57=45.7 moles

Moles of nitrogen = 45.7*0.79=36.1

Products contain nitrogen, excess oxygen, CO2 and H2O

Products ( dry basis ) : N2= 36.1 ,excess oxygen = 45.7*0.21-6=3.45 moles, CO2= 4 moles

Total moles of products : 36.1+3.45+4= 43.55 moles

Percentages : N2: 100*36.1/43.55=82.89% Oxygen= 100*3.45/43.55=7.92% and CO2= 100*4/43.55=9.18%

Products ( Wet basis ) : N2= 36.1 ,excess oxygen = 45.7*0.21-6=3.45 moles, CO2= 4 moles, H2O= 5 moles

Total moles of products : 36.1+3.45+4+5= 48.55 moles

Percentages : N2: 100*36.1/48.55=74.35% , O2= 100*3.45/48.5=7.11%, CO2= 100*4/48.55=8.24% and H2O= 100*5/48.55=10.29%

Total moles of products on dry basis : 43.55 moles

P= 100Kpa= 100/101.33 atm =0.98 atm, T= 260 deg.c =260+273.15=533.15K, R=0.08206L.atm/mole.K

Volume on dry basis , V= nRT/P, where n= number of moles= 43.55

Volume on dry basis = 43.55*0.08206*533.15/ 0.98=1994.21 L

On wet basis, volume = 48.55*0.08206*533.15/0.98=2167.425 L