Answer:

Orsat Analysis of Flue Gas

%CO₂ = 10.56%

%N₂ = 81.08%

%O₂ = 8.36%

Step-by-step explanation

On the basis of 100kmol fuel, we compute the moles of the following:

n_CO = 100kmol(0.50) = 50kmol

n_H2 = 100kmol (0.30) = 30kmol

n_C2H6 = 100kmol(0.05) = 5kmol

n_O2 = 100kmol(0.10) = 10kmol

Reactions based on complete combustion

2CO + O₂ →2CO₂

2C₂H₆ + 14O₂ → 4CO₂ + 3H₂0

The moles of CO and C₂H₆ converted to CO₂

n_CO2 = 50kmol CO (2 kmol CO2/ 2kmol CO) + (5kmol C₂H₆) (4 kmol CO2/ 2kmol CO)

n_CO2 = 60 kmol

Theoretical oxygen in order to attain complete combustion

n_{Theo O2} = 50 kmol ( 1 kmol O₂ / 2 kmol CO) + (30 kmol H₂) (1kmol O₂ / 2kmol H₂) + (5 kmol C₂H₆ (14kmol O2 / 2 kmol C₂H₆)

n_{Theo O2} =75 kmol

Moles of oxygen fed

n_O2total= 10 kmol O₂ + 75 kmol O₂ (1.5)

n_O2total= 122.5 kmol

Moles of unreacted O₂

n_O2 = n_O2total - n_{Theo O2}

n_O2 =122.5 kmol - 75 kmol

n_O2 = 47.5 kmol

Moles of nitrogen

n_N2 = 122.5 kmol O₂ (79 kmol N₂ / 21 kmol O₂)

n_N2 = 460.83 kmol

Total moles of flue gas

n_T = n_N2 + n_O2 +n_CO2

n_T = 460.83 kmol + 47.5 kmol + 60 kmol

n_T = 568.33 kmol

%CO₂ =(n_CO2 /n_T )(100%)

%CO₂ =(60 kmols / 568.33 kmol )(100%)

%CO₂ = 10.56%

%N₂ =(n_N2/n_T )(100%)

%N₂ =(460.83 kmol/ 568.33 kmol)(100%)

%N₂ = 81.08%

%O₂ = (nO2/n_T )(100%)

%O₂ = (47.5kmol/568.33kmol)(100%)

%O₂ = 8.36%