Answer to Question #106160 in General Chemistry for odusanya samson

Question

Ethane is burned with 50% excess air. the percentage conversionof the ethane is 90%. from the ethane that is converted , 25% react to form CO and the rest react to form CARBONDIOXIDE.
what is the molar composition of the stack gas on a dry basis? what is the mole ratio of water to dry stack gas

Answer 1

n₀ - initial

n - final

2C₂​H₆​+7O₂​→4CO₂​+6H_2​O

2C₂​H_6​+5O₂​→4CO+6H_2​O

Let n₀(C₂​H_6​) be 1 mol then

n₀(O₂​) = 1/2*7*1.5 = 5.25 mol

n(C₂​H₆​) = 0.1 mol

n(O₂) = 5.25 - 1*0.9*0.25/2*5 - 1*0.9*0.75/2*7 = 2.325 mol

n(CO₂) = 1*0.9*0.75/2*4 = 1.35 mol

n(CO) = 1*0.9*0.25/2*4 = 0.45 mol

n(H₂O) = 1*0.9*0.75/2*6 + 1*0.9*0.25/2*6 = 2.7 mol

the molar composition of the stack gas on a dry basis:

C₂​H₆ 0.1/(0.1+2.325+1.35+0.45)*100 = 2.4%

O₂ 2.325/(0.1+2.325+1.35+0.45)*100 = 55.0%

CO₂ 1.35/(0.1+2.325+1.35+0.45)*100 = 32.0%

CO 0.45/(0.1+2.325+1.35+0.45)*100 = 10.6%

the mole ratio of water to dry stack gas:

2.7/(0.1+2.325+1.35+0.45) = 0.639