Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 .\) L/min at \(1.50\) atm and ambient temperature. Air is added to the chamber at 1.00 atm and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is \(21\) mole percent \(\mathrm{O}_{2}\) and \(79\) mole percent \(\mathrm{N}_{2}\), calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that \(95.0 \%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\). The remainder was present as carbon in \(\mathrm{CO}\). Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.

Step 1: Write the balanced chemical equation for complete combustion of methane \[CH_4 (g) + 2O_2 (g) -> CO_2 (g) + 2H_2O (g)\] Step 2: Determine the moles of oxygen required for complete combustion of methane Given that methane is flowing at 200 L/min at 1.50 atm and ambient temperature, we can use the ideal gas law to find the moles of methane flowing per minute. \(PV=nRT\) n = PV/RT Assume the ambient temperature is 298 K, and take the gas constant R = 0.0821 L.atm/mol.K. Methane moles (n\(CH_4\))= (200 L/min)(1.50 atm) / ((0.0821 L.atm/mol.K)(298 K)) ≈ 12.19 mol/min Using the stoichiometric ratio of the balanced equation above (1 mol\(CH_4\): 2 mol\(O_2\)): Moles of oxygen required (n\(O_2\)) = 12.19 mol/min * 2 = 24.38 mol/min Step 3: Calculate the flow rate of air necessary to deliver the required amount of oxygen Three times the amount of oxygen is necessary, therefore, n\(O_2\) required in air = 24.38 * 3 = 73.14 mol/min Since air is 21% oxygen, we can find the total moles of air needed per minute. Air moles (n\(air\)) = 73.14 / 0.21 ≈ 348.28 mol/min Now use the ideal gas law again to find the flow rate of air at 1.00 atm and 298 K: Flow rate of air (V\(air\))= n\(air\)RT/P ≈ (348.28 mol/min)(0.0821 L.atm/mol.K)(298 K)/(1.00 atm) ≈ 8524 L/min The flow rate of air necessary to deliver the required amount of oxygen is approximately 8524 L/min.

Step 1: Calculate the moles of CO and CO₂ in the exhaust gas 95% of carbon in the exhaust gas is present as CO₂, remaining 5% as CO. Since 1 mol of CH₄ produces 1 mol of carbon: n\(CO_2\) = 0.95 * 12.19 = 11.58 mol/min n\(CO\) = 0.05 * 12.19 = 0.61 mol/min Step 2: Calculate the moles of unreacted O₂ and unreacted N₂ in the exhaust gas Moles of O₂ consumed in the formation of CO₂ and CO = 2 * 12.19 = 24.38 mol/min 3 * n\(O_2\) = 73.14 mol/min is present in the air, therefore, the unreacted moles of O₂ are: Unreacted O₂ = 73.14 - 24.38 = 48.76 mol/min Since air is 79% N₂, unreacted \(N_2\) = 0.79 * 348.28 = 275.14 mol/min Step 3: Calculate the moles of produced H₂O From the balanced equation, 1 mol of CH₄ produces 2 mol of H₂O: n\(H_2O\) = 2 * 12.19 = 24.38 mol/min Step 4: Calculate the mole fractions of the exhaust gas components Total moles in the exhaust gas = n\(CO_2\) + n\(CO\) + unreacted O₂ + unreacted N₂ + n\(H_2O\) Total moles = 11.58 + 0.61 + 48.76 + 275.14 + 24.38 = 360.47 mol/min Mole fractions of each component in the exhaust gas: X\(CO_2\) = n\(CO_2\)/total moles = 11.58 / 360.47 ≈ 0.0321 X\(CO\) = n\(CO\)/total moles = 0.61 / 360.47 ≈ 0.0017 X\(O_2\) = unreacted O₂/total moles = 48.76 / 360.47 ≈ 0.1353 X\(N_2\) = unreacted N₂/total moles = 275.14 / 360.47 ≈ 0.7631 X\(H_2O\) = n\(H_2O\)/total moles = 24.38 / 360.47 ≈ 0.0677 The composition of the exhaust gas in terms of mole fractions is approximately CO: 0.0017, CO₂: 0.0321, O₂: 0.1353, N₂: 0.7631 and H₂O: 0.0677