i (20 pts) A burner is fueled with methane (CH4) The analysis of the exhaust gas

Question

i (20 pts) A burner is fueled with methane (CH4) The analysis of the exhaust gas is: N2 72.88 mole percent 02 3.87 mole percent CO27.75 mole percent H2Q 15.5 mole percent Total 100.00 mole percent Please answer the following questions: a. (5 pts) Show the balanced chemical reaction depicting complete combustion of this fuel. b. (5 pts) For 100 moles of exhaust gas, how many moles of methane are burned? c. (5 pts) Assume air is 79% N, and 2100 O. For 100 moles of exhaust gas, how many moles of air was fed to the burner? d. (5 pts) What is the % of excess air?

Explanation / Answer

Part a

Balanced equation

CH4 + 2(O2 + 3.76N2) = CO2 + 2H2O + 7.52 N2

Part b

In exhaust gas

Total Moles = 100 mol

Moles of CO2 = 100 x 7.75 = 7.75 mol

From the stoichiometry of the reaction

1 mol CO2 produced from = 1 mol CH4

7.75 mol CO2 produced from = 7.75 mol CH4

Moles of methane burned = 7.75 mol

Part c

From the stoichiometry of the reaction

1 mol CH4 required = 2 mol O2

7.75 mol CH4 required = 2 x 7.75 = 15.5 mol O2

Moles of O2 consumed = 15.5 mol

Moles of O2 fed = Moles of O2 consumed + moles of O2 in exhaust gas

= 15.5 + 3.87

= 19.37 mol

Air consists 21 mol % of O2 and 79 mol % of N2

Moles of air fed = 19.37 mol / 0.21 = 92.25 mol

Moles of N2 fed = 92.24 x 0.79 = 72.88 mol

Moles of N2 in exhaust gas = 72.88 mol

Our calculation is perfect.

Part d

From the stoichiometry of the reaction

1 mol O2 required = 2 + 7.52 = 9.52 mol air

7.75 mol CH4 required = 7.75 x 9.52 = 73.8 mol air

Moles of air in excess = moles of air fed - moles of air reacted

= 92.25 - 73.8

= 18.45 mol

% of excess air = moles of air in excess x 100 / moles of air reacted

= (18.45*100) / 73.8

= 25 %

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