Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of 200\. \(\mathrm{L} / \mathrm{min}\) at \(1.50 \mathrm{~atm}\) and ambient temperature. Air is added to the chamber at \(1.00 \mathrm{~atm}\) and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(\mathrm{~g})\) 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 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.

For complete combustion of methane to produce carbon dioxide and water vapor, the balanced chemical equation is: \[ \mathrm{CH}_4 + 2\ \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\ \mathrm{H}_2\mathrm{O} \]

Using the ideal gas law, we can find the molar flow rate of methane. The ideal gas law is given by \(PV=nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the moles, \(R\) is the gas constant, and \(T\) is the temperature. Here, \(P = 1.50\ \mathrm{atm}\), \(V = 200\ \mathrm{L/min}\), and \(T\) is ambient temperature. To find the molar flow rate of methane, we need to rearrange the ideal gas law as follows: \[n_{\mathrm{CH_4}} = \frac{PV}{RT}\] Assuming the ambient temperature to be 298 K (25°C) and using the appropriate gas constant (\(0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}\)), we get: \[n_{\mathrm{CH_4}} = \frac{1.50\ \mathrm{atm} \times 200\ \mathrm{L/min}}{(0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}})(298\ \mathrm{K})}\] \[n_{\mathrm{CH_4}} = 12.2\ \mathrm{mol/min}\]

Using the stoichiometry of the balanced chemical equation, 2 moles of oxygen are required for every mol of methane. So, the molar flow rate of oxygen required (\(n_{\mathrm{O_2, required}}\)) is: \[n_{\mathrm{O_2, required}} = 2 \times n_{\mathrm{CH_4}}\] \[n_{\mathrm{O_2, required}} = 2 \times 12.2\ \mathrm{mol/min}\] \[n_{\mathrm{O_2, required}} = 24.4\ \mathrm{mol/min}\]

It is given that three times as much oxygen as necessary is reacted. Therefore, the molar flow rate of oxygen supplied (\(n_{\mathrm{O_2, supplied}}\)) is: \[n_{\mathrm{O_2, supplied}} = 3 \times n_{\mathrm{O_2, required}}\] \[n_{\mathrm{O_2, supplied}} = 3 \times 24.4\ \mathrm{mol/min}\] \[n_{\mathrm{O_2, supplied}} = 73.2\ \mathrm{mol/min}\]

It is given that the air is 21 mole percent oxygen and 79 mole percent nitrogen. So, the flow rate of air (\(V_{air}\)) necessary to deliver the required amount of oxygen can be calculated by rearranging the flow rate of oxygen supplied as follows: \[V_{air} = \frac{n_{\mathrm{O_2, supplied}}}{0.21}\] \[V_{air} = \frac{73.2\ \mathrm{mol/min}}{0.21}\] \[V_{air} = 348.6\ \mathrm{mol/min}\] b. Calculating the composition of the exhaust gas under incomplete combustion

It is given that 95% of the carbon in the exhaust gas is present in CO2 and the remainder is present as carbon in CO. Therefore, for every mol of methane reacted, 0.95 mol CO2 and 0.05 mol CO are produced.

We will first determine the molar flow rates of CO2, CO, O2, N2, and H2O in the exhaust gas using the stoichiometry of the reactions. \[ n_{\mathrm{CO_2}} = 0.95 \times n_{\mathrm{CH_4}} \] \[ n_{\mathrm{CO_2}} = 0.95 \times 12.2\ \mathrm{mol/min} \] \[ n_{\mathrm{CO_2}} = 11.6\ \mathrm{mol/min} \] Similarly, for CO: \[ n_{\mathrm{CO}} = 0.05 \times n_{\mathrm{CH_4}} \] \[ n_{\mathrm{CO}} = 0.05 \times 12.2\ \mathrm{mol/min} \] \[ n_{\mathrm{CO}} = 0.61\ \mathrm{mol/min} \] For O2, we know that the molar flow rate of oxygen supplied is 73.2 mol/min and 24.4 mol/min is consumed in the reaction, so the remaining oxygen in the exhaust gas is: \[ n_{\mathrm{O_2}} = 73.2\ \mathrm{mol/min} - 24.4\ \mathrm{mol/min} \] \[ n_{\mathrm{O_2}} = 48.8\ \mathrm{mol/min} \] Now calculate the molar flow rate of nitrogen (N2) in the exhaust gas: \[n_{\mathrm{N_2}} = 0.79 \times V_{air}\] \[n_{\mathrm{N_2}} = 0.79 \times 348.6\ \mathrm{mol/min}\] \[n_{\mathrm{N_2}} = 275.4\ \mathrm{mol/min}\] For H2O, we know that 2 mol of H2O is produced for every mol of methane, so: \[ n_{\mathrm{H_2O}} = 2 \times n_{\mathrm{CH_4}} \] \[ n_{\mathrm{H_2O}} = 2 \times 12.2\ \mathrm{mol/min} \] \[ n_{\mathrm{H_2O}} = 24.4\ \mathrm{mol/min} \]

Find the total moles in the exhaust gas: \[n_{total} = n_{\mathrm{CO_2}} + n_{\mathrm{CO}} + n_{\mathrm{O_2}} + n_{\mathrm{N_2}} + n_{\mathrm{H_2O}}\] \[n_{total} = 11.6\ \mathrm{mol/min} + 0.61\ \mathrm{mol/min} + 48.8\ \mathrm{mol/min} + 275.4\ \mathrm{mol/min} + 24.4\ \mathrm{mol/min}\] \[n_{total} = 360.8\ \mathrm{mol/min}\] Finally, calculate the mole fractions of each component in the exhaust gas: \[y_{\mathrm{CO_2}} = \frac{n_{\mathrm{CO_2}}}{n_{total}}\] \[y_{\mathrm{CO_2}} = \frac{11.6\ \mathrm{mol/min}}{360.8\ \mathrm{mol/min}}\] \[y_{\mathrm{CO_2}} = 0.0322\] Similarly, \[y_{\mathrm{CO}} = \frac{n_{\mathrm{CO}}}{n_{total}}\] \[y_{\mathrm{CO}} = \frac{0.61\ \mathrm{mol/min}}{360.8\ \mathrm{mol/min}}\] \[y_{\mathrm{CO}} = 0.0017\] \[y_{\mathrm{O_2}} = \frac{n_{\mathrm{O_2}}}{n_{total}}\] \[y_{\mathrm{O_2}} = \frac{48.8\ \mathrm{mol/min}}{360.8\ \mathrm{mol/min}}\] \[y_{\mathrm{O_2}} = 0.1353\] \[y_{\mathrm{N_2}} = \frac{n_{\mathrm{N_2}}}{n_{total}}\] \[y_{\mathrm{N_2}} = \frac{275.4\ \mathrm{mol/min}}{360.8\ \mathrm{mol/min}}\] \[y_{\mathrm{N_2}} = 0.7632\] \[y_{\mathrm{H_2O}} = \frac{n_{\mathrm{H_2O}}}{n_{total}}\] \[y_{\mathrm{H_2O}} = \frac{24.4\ \mathrm{mol/min}}{360.8\ \mathrm{mol/min}}\] \[y_{\mathrm{H_2O}} = 0.0676\] Thus, the composition of the exhaust gas in terms of mole fraction of CO, CO2, O2, N2, and H2O is approximately 0.0017, 0.0322, 0.1353, 0.7632, and 0.0676, respectively.