Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of 200. L/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 \(79 \mathrm{~mole}\) 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.

Given the flow rate of methane at 200 L/min and the pressure and temperature being 1.5 atm and ambient, respectively, we can use the ideal gas law to find the moles of methane flowing per minute: \[PV=nRT\] \[n=\frac{PV}{RT}\] Let's use the value of R = 0.0821 L atm/mol K. Assuming the ambient temperature to be 298 K, we can now calculate the moles of methane per minute: \[n_{\mathrm{CH}_{4}}=\frac{(1.5\,\mathrm{atm})(200\,\mathrm{L/min})}{(0.0821\,\mathrm{L\,atm/mol\,K})(298\,\mathrm{K})} = 12.15\,\mathrm{mol/min}\]

To ensure complete combustion of methane, the balanced chemical equation is: \[\mathrm{CH}_{4}+\mathrm{2O}_{2}\rightarrow \mathrm{CO}_{2}+\mathrm{2H}_{2}\mathrm{O}\] According to the equation, one mole of methane requires two moles of oxygen. Since three times the required amount is reacted, the moles of oxygen required per minute are: \[n_{\mathrm{O}_{2}}=12.15\,\mathrm{mol/min}\times 2\times 3=72.90\,\mathrm{mol/min}\]

Air is 21 mole percent \(\mathrm{O}_{2}\), so the amount of air required to deliver the moles of oxygen is: \[\mathrm{Air\,flow\,rate}=\frac{72.90\,\mathrm{mol/min}}{0.21}=347.1\,\mathrm{mol/min}\]

Since \(95.0\%\) of carbon is present in the form of \(\mathrm{CO}_{2}\), we can calculate the moles of \[\mathrm{CO}_{2}\] and \[\mathrm{CO}\] produced using the conversion as follows: \[n_{\mathrm{CO}_{2}}=12.15\,\mathrm{mol/min}\times 0.95 = 11.54\,\mathrm{mol/min}\] \[n_{\mathrm{CO}}=12.15\,\mathrm{mol/min}\times 0.05 = 0.608\,\mathrm{mol/min}\] Calculate the moles of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) left unreacted: \[n_{\mathrm{O}_{2,unreacted}}=347.1\,\mathrm{mol/min}\times 0.21 - 72.90\,\mathrm{mol/min} = 1.47\,\mathrm{mol/min}\] \[n_{\mathrm{N}_{2,unreacted}}=347.1\,\mathrm{mol/min}\times 0.79 = 274.2\,\mathrm{mol/min}\] Calculate the moles of \(\mathrm{H}_{2} \mathrm{O}\) produced: \[n_{\mathrm{H}_{2} \mathrm{O}}=12.15\,\mathrm{mol/min}\times 2 = 24.3\,\mathrm{mol/min}\] Calculate the mole fraction of each component: \[x_{\mathrm{CO}}=\frac{0.608\,\mathrm{mol/min}}{312.1\,\mathrm{mol/min}}=0.00195\] \[x_{\mathrm{CO}_{2}}=\frac{11.54\,\mathrm{mol/min}}{312.1\,\mathrm{mol/min}}=0.0369\] \[x_{\mathrm{O}_{2}}=\frac{1.47\,\mathrm{mol/min}}{312.1\,\mathrm{mol/min}}=0.00471\] \[x_{\mathrm{N}_{2}}=\frac{274.2\,\mathrm{mol/min}}{312.1\,\mathrm{mol/min}}=0.879\] \[x_{\mathrm{H}_{2} \mathrm{O}}=\frac{24.3\,\mathrm{mol/min}}{312.1\,\mathrm{mol/min}}=0.0779\] So, the exhaust gas composition is: - \(\mathrm{CO}: 0.00195\) - \(\mathrm{CO}_{2}: 0.0369\) - \(\mathrm{O}_{2}: 0.00471\) - \(\mathrm{N}_{2}: 0.879\) - \(\mathrm{H}_{2} \mathrm{O}: 0.0779\)