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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 Oâ‚‚, we can calculate the flow rate of the air by dividing the total number of moles of Oâ‚‚ 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 COâ‚‚ in the exhaust gas

Since 95% of carbon in the exhaust gas is present in COâ‚‚ and the rest is present in CO, we can find the number of moles of CO and COâ‚‚ 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 Oâ‚‚ and Nâ‚‚ in the exhaust gas

Since the flow rate of Oâ‚‚ entering the chamber is three times the required amount and all CHâ‚„ reacts: \(n_{O_2}^{unreacted} = n_{O_2}^{total} - 2 * n_{CH4} = 73.50 - 24.50 = 49.00\, \text{moles/min}\) The amount of Nâ‚‚ 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 Hâ‚‚O in the exhaust gas

According to the balanced equation, 2 moles of Hâ‚‚O are produced for each mole of CHâ‚„ 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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law

The Ideal Gas Law is an equation of state for a fictitious ideal gas. Even though real gases deviate from the ideal behavior, this equation provides a useful approximation for gases under many conditions. The formula is expressed as:

• \( PV = nRT \)

Here, \( P \) represents pressure, \( V \) is volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is the absolute temperature measured in Kelvin.
To apply the Ideal Gas Law, ensure that:

• Pressure is measured in atm (atmospheres).
• Volume is measured in liters (L).
• The number of moles (n) represents the quantity of gas molecules.
• Temperature (T) is in Kelvin (K), which you can convert from Celsius by adding 273.15.

This law allows us to convert between the volume of a gas and the number of moles it contains when variables like pressure and temperature are constant, as with methane in the given problem.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It involves using a balanced chemical equation to relate the masses, moles, and volumes of reaction constituents. For example, in combusting methane:

• \( \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g) \)

This equation shows that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.
In practice, stoichiometry helps calculate the amount of reactants needed or products formed in a reaction. By understanding the molar ratios, you can determine the amount of oxygen required for the complete combustion of a given quantity of methane.

Mole Fraction

Mole fraction is a way to express the concentration of a component in a mixture. It represents the ratio of the number of moles of one component to the total number of moles in the mixture. The mole fraction (\( X \)) can be calculated using the formula:

• \( X_i = \frac{n_i}{n_{total}} \)

Where \( n_i \) is the number of moles of component \( i \), and \( n_{total} \) is the total number of moles in the mixture.
In the methane combustion problem, you need to determine the mole fractions of different gases in the exhaust, such as CO, COâ‚‚, Oâ‚‚, Nâ‚‚, and Hâ‚‚O. Calculating mole fractions helps in understanding composition in terms of concentration, which is critical for evaluating the results of chemical reactions.

Exhaust Gas Composition

The composition of exhaust gases provides insight into the outcome of combustion reactions. These include not just the desired products but also any remaining reactants or by-products. Incomplete combustion can result in the presence of not only carbon dioxide but also carbon monoxide.
Understanding the exhaust gas composition involves determining:

• The amounts of reacted and unreacted gases, like the remaining Oâ‚‚.
• By-products formed, such as CO when complete combustion doesn't occur.
• The ratios or mole fractions of all gases in the exhaust mixture.

Evaluating exhaust gases is crucial for determining efficiency and safety of combustion processes. This involves ensuring that the proportion of each gas remains within acceptable limits, which helps in minimizing pollutants like CO.