Question

Natural gas consists primarily of methane, \(\mathrm{CH}_{4}(g) .\) (a) Write a balanced chemical equation for the complete combustion of methane to produce \(\mathrm{CO}_{2}(g)\) as the only carbon-containing product. (b) Write a balanced chemical equation for the incomplete combustion of methane to produce \(\mathrm{CO}(g)\) as the only carbon-containing product. (c) At \(25^{\circ} \mathrm{C}\) and 1.0 atm pressure, what is the minimum quantity of dry air needed to combust \(1.0 \mathrm{~L}\) of \(\mathrm{CH}_{4}(g)\) completely to \(\mathrm{CO}_{2}(g) ?\)

Short answer

(a) The balanced chemical equation for the complete combustion of methane is: \[\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)}\] (b) The balanced chemical equation for the incomplete combustion of methane is: \[\mathrm{CH_4(g) + 3/2O_2(g) \rightarrow CO(g) + 2H_2O(g)}\] (c) At 25°C and 1.0 atm pressure, the minimum quantity of dry air needed to combust 1.0 L of methane gas completely to carbon dioxide gas is 0.3887 L.

Step-by-step solution

Complete Combustion Equation

The complete combustion of methane (\(\mathrm{CH_4}\)) involves the reaction between methane and oxygen (\(\mathrm{O_2}\)), producing carbon dioxide (\(\mathrm{CO_2}\)) and water (\(\mathrm{H_2O}\)). The balanced chemical equation for this reaction is: \[\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)}\] (b) Incomplete Combustion of Methane

Incomplete Combustion Equation

The incomplete combustion of methane (\(\mathrm{CH_4}\)) occurs when there is not enough oxygen to create CO2 as a product. In such cases, the products include carbon monoxide (\(\mathrm{CO}\)) and water (\(\mathrm{H_2O}\)). The balanced chemical equation for this reaction is: \[\mathrm{CH_4(g) + 3/2O_2(g) \rightarrow CO(g) + 2H_2O(g)}\] (c) Calculation of Minimum Dry Air

Molar Volume of an Ideal Gas

At 25°C and 1.0 atm pressure, we can use the molar volume of an ideal gas, which is approximately 24.5 L/mol. In other words, one mole of gas occupies a volume of 24.5 L under these conditions.

Moles of Methane and Oxygen Needed

We have 1.0 L of methane gas, so we can calculate the number of moles of methane in this volume: \[\frac{1.0\,\text{L}\,\mathrm{CH_4}}{24.5\,\text{L}\,/\,\text{mol}} = 0.04082\,\text{mol}\,\mathrm{CH_4}\] Looking at the balanced equation for complete combustion, we see that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide. Therefore, to completely combust 0.04082 moles of methane, we will need: \[(0.04082\,\text{mol}\,\mathrm{CH_4})\left(\frac{2\,\text{mol}\,\mathrm{O_2}}{1\,\text{mol}\,\mathrm{CH_4}}\right) = 0.08163\,\text{mol}\,\mathrm{O_2}\]

Composition of Dry Air and Volume of Air Needed

Dry air is composed of approximately 21% oxygen by volume. To calculate the minimum quantity of dry air needed to provide 0.08163 moles of oxygen, we can use the following proportion: \[\frac{0.08163\,\text{mol}\,\mathrm{O_2}}{x\,\text{L}\,\text{air}} = \frac{0.21\,\text{mol}\,\mathrm{O_2}}{1\,\text{L}\,\text{air}}\] Now we solve for x: \[x = \frac{0.08163\,\text{mol}\,\mathrm{O_2}}{0.21\,\text{mol}\,\mathrm{O_2}\,/\,\text{L}\,\text{air}} = 0.3887\,\text{L}\,\text{air}\]

Conclusion

At 25°C and 1.0 atm pressure, the minimum quantity of dry air needed to combust 1.0 L of methane gas completely to carbon dioxide gas is 0.3887 L.

Key concepts

Balanced Chemical Equations

In chemistry, a balanced chemical equation is a representation that shows the equality of atoms from reactants to products during a chemical reaction.
It is crucial because it adheres to the law of conservation of mass, ensuring every atom is accounted for throughout the process. For example, in the complete combustion of methane, the balanced equation is \[\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)}\]This equation indicates that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.

• Reactants: 1 methane molecule, 2 oxygen molecules.
• Products: 1 carbon dioxide molecule, 2 water molecules.

Balancing equations ensures that equal numbers of each type of atom are present on both the reactant and product sides of the equation, which is fundamental for correctly describing chemical reactions.

Methane Combustion

Methane combustion occurs when methane, a simple hydrocarbon (\(\mathrm{CH_4}\)) reacts with oxygen (\(\mathrm{O_2}\)), resulting in the formation of different products depending on the oxygen availability.
Let's explore its two forms:- **Complete Combustion**: Happens with an ample supply of oxygen, forming carbon dioxide (\(\mathrm{CO_2}\)) and water (\(\mathrm{H_2O}\)). The reaction is exothermic, releasing energy, as given by the equation:\[\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)}\]- **Incomplete Combustion**: Occurs with insufficient oxygen, leading to the production of carbon monoxide (\(\mathrm{CO}\)) and water (\(\mathrm{H_2O}\)). This form is not desirable as carbon monoxide is a dangerous pollutant. The equation is:\[\mathrm{CH_4(g) + \frac{3}{2}O_2(g) \rightarrow CO(g) + 2H_2O(g)}\]Methane combustion is a key reaction in energy production due to its ability to release substantial heat, making it indispensable in various industrial processes.

Ideal Gas Law

The Ideal Gas Law is a vital principle in chemistry that relates the four variables of a gas: pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and number of moles (\(n\)).
The formula is expressed as:\[PV = nRT\]where \(R\) is the gas constant.This law allows chemists to predict the behavior of a gas under various conditions and is instrumental in calculating quantities in gas reactions.

Application in Methane Combustion

In the problem of methane combustion, the Ideal Gas Law is used to find the number of moles of methane:Given 1.0 L of methane at 25°C and 1 atm, known conditions are simplified using the molar volume of a gas at standard conditions:

• Volume Used: 1.0 L
• Moles of Methane: Calculated as \(\frac{1.0 \, \text{L}}{24.5 \, \text{L/mol}} = 0.04082 \, \text{mol}\)

This approach helps in further calculations involving the moles of oxygen needed for complete combustion, underscoring how the Ideal Gas Law facilitates understanding and predicting gas behavior.

Stoichiometry

Stoichiometry is the weighing and measuring of chemical elements and reactions, offering an essential method for calculating reactant and product quantities.
In chemistry, stoichiometry provides a quantitative basis that is crucial for understanding reaction mechanisms and yields.

Importance in Combustion

For the combustion of methane, stoichiometry helps determine how much oxygen is needed to fully react with a given amount of methane.Given the balanced equation for complete combustion:\[\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)}\]

• 1 mole of methane reacts with 2 moles of oxygen.
• From the problem, there are 0.04082 moles of methane.

Using stoichiometry, we calculate the required moles of oxygen:\[0.04082 \, \text{mol} \, \text{CH}_4 \times \frac{2 \, \text{mol} \, \text{O}_2}{1 \, \text{mol} \, \text{CH}_4} = 0.08163 \, \text{mol} \, \text{O}_2\]This calculation is essential to ensure complete combustion, and stoichiometry makes it possible to determine such ratios with precision, deeply connected to the balanced chemical equation and reaction conditions.