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 \(\mathrm{CO}_{2}(g)\) 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 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 by step solution

01

Determine the molar flow rate of methane

We're given that the flow rate of methane is 200 L/min at 1.50 atm and ambient temperature. To convert from volume to moles, we can use the ideal gas law: PV = nRT First, we need to express the temperature in Kelvin. Assuming an ambient temperature of 25°C, we have: T = 25 + 273.15 = 298.15 K Rearranging the ideal gas law to find the molar flow rate (n/min), we get: \(n = \frac{P * V}{R * T}\) For methane, we have: P = 1.50 atm V = 200 L/min R = 0.0821 L * atm / K * mol T = 298.15 K Plugging in the values: \(n_{CH4} = \frac{1.50 * 200}{0.0821 * 298.15} = 12.25\, \text{moles/min}\)

02

Determine the molar flow rate of oxygen required

The balanced equation for the complete combustion of methane is: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) According to the stoichiometry, 2 moles of O₂ are required for the complete combustion of 1 mole of CH₄. Therefore, to determine the required flow rate of oxygen, we multiply the molar flow rate of methane by 2: \(n_{O_2}^{required} = 2 * n_{CH4} = 2 * 12.25 = 24.50\, \text{moles/min}\)

03

Calculate the flow rate of air

The exercise states that three times the required amount of oxygen is needed. We find the total amount of oxygen entering the combustion chamber: \(n_{O_2}^{total} = 3 * n_{O_2}^{required} = 3 * 24.50 = 73.50\, \text{moles/min}\) Since air is 21 mole percent O₂, we can calculate the flow rate of the air by dividing the total number of moles of O₂ by the mole percentage: \(n_{air} = \frac{n_{O_2}^{total}}{0.21} = \frac{73.50}{0.21} = 350.00\, \text{moles/min}\)

04

Convert the molar flow rate of air to volume flow rate

We can use the ideal gas law again to convert from moles to volume: PV = nRT Rearranging for volume: V = n * R * T / P For air: P = 1.00 atm Plugging in the values: \(V_{air} = \frac{350.00 * 0.0821 * 298.15}{1.00} = 8576 \, \text{L/min}\) The flow rate of air necessary is 8576 L/min. #b. Composition of the exhaust gas#

05

Find the amounts of CO and CO₂ in the exhaust gas

Since 95% of carbon in the exhaust gas is present in CO₂ and the rest is present in CO, we can find the number of moles of CO and CO₂ produced: \(n_{CO_2} = 0.95 * n_{CH4} = 0.95 * 12.25 = 11.64\, \text{moles/min}\) \(n_{CO} = 0.05 * n_{CH4} = 0.05 * 12.25 = 0.613\, \text{moles/min}\)

06

Determine the amounts of unreacted O₂ and N₂ in the exhaust gas

Since the flow rate of O₂ entering the chamber is three times the required amount and all CH₄ reacts: \(n_{O_2}^{unreacted} = n_{O_2}^{total} - 2 * n_{CH4} = 73.50 - 24.50 = 49.00\, \text{moles/min}\) The amount of N₂ in the exhaust gas remains unchanged since it doesn't react: \(n_{N2} = 0.79 * n_{air} = 0.79 * 350.00 = 276.5\, \text{moles/min}\)

07

Calculate the amount of H₂O in the exhaust gas

According to the balanced equation, 2 moles of H₂O are produced for each mole of CH₄ that reacts: \(n_{H_2O} = 2 * n_{CH4} = 2 * 12.25 = 24.50\, \text{moles/min}\)

08

Determine the mole fractions of the exhaust gas components

To find the mole fractions, we divide the number of moles of each component by the total number of moles in the exhaust gas: \(X_{CO} = \frac{n_{CO}}{n_{CO} + n_{CO_2} + n_{O_2}^{unreacted} + n_{N2} + n_{H_2O}} = \frac{0.613}{0.613 + 11.64 + 49.00 + 276.5 + 24.50} = 0.00172\) \(X_{CO_2} = \frac{n_{CO_2}}{n_{CO} + n_{CO_2} + n_{O_2}^{unreacted} + n_{N2} + n_{H_2O}} = \frac{11.64}{0.613 + 11.64 + 49.00 + 276.5 + 24.50} = 0.04107\) \(X_{O_2} = \frac{n_{O_2}^{unreacted}}{n_{CO} + n_{CO_2} + n_{O_2}^{unreacted} + n_{N2} + n_{H_2O}} = \frac{49.00}{0.613 + 11.64 + 49.00 + 276.5 + 24.50} = 0.17325\) \(X_{N_2} = \frac{n_{N2}}{n_{CO} + n_{CO_2} + n_{O_2}^{unreacted} + n_{N2} + n_{H_2O}} = \frac{276.5}{0.613 + 11.64 + 49.00 + 276.5 + 24.50} = 0.77600\) \(X_{H_2O} = \frac{n_{H_2O}}{n_{CO} + n_{CO_2} + n_{O_2}^{unreacted} + n_{N2} + n_{H_2O}} = \frac{24.50}{0.613 + 11.64 + 49.00 + 276.5 + 24.50} = 0.08667\) The composition of the exhaust gas in terms of mole fractions is: \(X_{CO} = 0.00172\), \(X_{CO_2} = 0.04107\), \(X_{O_2} = 0.17325\), \(X_{N_2} = 0.77600\), and \(X_{H_2O} = 0.08667\).

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