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 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 \(\mathrm{CH}_{4}(\mathrm{g})\) completely to \(\mathrm{CO}_{2}(\mathrm{g}) ?\)

Short answer

(a) The balanced chemical equation for the complete combustion of methane is: \( CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g) \) (b) The balanced chemical equation for the incomplete combustion of methane is: \( CH_{4}(g) + 3/2O_{2}(g) \rightarrow CO(g) + 2H_{2}O(g) \) (c) The minimum quantity of dry air needed to combust 1.0 L of CH4(g) completely to CO2(g) at 25°C and 1.0 atm pressure is approximately 9.5 L.

Step-by-step solution

(a) Balanced Chemical Equation for Complete Combustion

To write the balanced chemical equation for the complete combustion of methane, we will combine methane (CH4) with the oxygen (O2) required for combustion, to produce carbon dioxide (CO2) and water (H2O) as products. The balanced equation is: \( CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g) \)

(b) Balanced Chemical Equation for Incomplete Combustion

For the incomplete combustion of methane, we will produce carbon monoxide (CO) as the only carbon-containing product, along with water (H2O). The balanced equation is: \( CH_{4}(g) + 3/2O_{2}(g) \rightarrow CO(g) + 2H_{2}O(g) \)

(c) Calculate Minimum Quantity of Dry Air

First, we need to find the minimum amount of oxygen required to combust 1.0 L of CH4(g). According to the balanced equation for complete combustion, 1 mole of CH4 reacts with 2 moles of O2. At 25°C and 1.0 atm pressure, 1 mole of any ideal gas occupies 24.45 L volume. Therefore, the amount of methane in moles is: \(n_{CH_{4}} = \frac{1.0 \, L}{24.45 \, L/mol} \approx 0.041 \, mol \) Now we can find the amount of oxygen required: \( n_{O_{2}} = 2 \times n_{CH_{4}} = 2 \times 0.041 \, mol = 0.082 \, mol \) The volume of oxygen required is: \( V_{O_{2}} = n_{O_{2}} \times 24.45\, L/mol = 0.082 \, mol \times 24.45\, L/mol \approx 2.0\, L \) Dry air is approximately 21% oxygen, so the minimum volume of dry air needed is: \( V_{Air} = \frac{V_{O_{2}}}{0.21} \approx \frac{2.0\, L}{0.21} \approx 9.5 \, L \) So, the minimum quantity of dry air needed to combust 1.0 L of CH4(g) completely to CO2(g) at 25°C and 1.0 atm pressure is approximately 9.5 L.

Key concepts

Chemical Equations

Chemical equations are a way to represent chemical reactions. They show how different substances interact, being transformed into new products. It's like a recipe that tells you what ingredients are needed and the final dish you'll get.

Substances in a chemical equation are represented by their chemical formulas, like \(\text{CH}_4\) for methane. Reactants are on the left side, the ingredients of our reaction, while the products, the new substances formed, are on the right side. An arrow separates them.

To balance a chemical equation, both sides of the equation must have the same number of each type of atom. This is due to the law of conservation of mass, which says that mass cannot be created or destroyed in a chemical reaction. For instance, when methane burns in oxygen to form carbon dioxide and water, the equation must reflect equal numbers of carbon, hydrogen, and oxygen atoms on both sides.

• Balanced equation for complete combustion of methane: \(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\)
• Balanced equation for incomplete combustion: \(\text{CH}_4 + \frac{3}{2}\text{O}_2 \rightarrow \text{CO} + 2\text{H}_2\text{O}\)

Stoichiometry

Stoichiometry is like the cookbook of chemistry. It lets us calculate the amounts of substances involved in chemical reactions. By using the coefficients from balanced chemical equations, stoichiometry helps determine how much of each reactant is needed, or how much of a product will be formed.

In our methane combustion, stoichiometry tells us that each molecule of methane requires two molecules of oxygen for complete combustion. When given an initial amount of one reactant, stoichiometry helps calculate how much of another reactant is necessary. For instance, if you have 1 mole of methane, you'll need 2 moles of oxygen for complete combustion, based on the balanced equation:

• 1 molecule \(\text{CH}_4\) + 2 molecules \(\text{O}_2\) = 1 molecule \(\text{CO}_2\) + 2 molecules \(\text{H}_2\text{O}\)
• Therefore, \(n_{\text{O}_2} = 2 \times n_{\text{CH}_4}\)

Methane Combustion

Methane combustion is a type of chemical reaction where methane gas reacts with oxygen to produce energy, typically as heat and light. This reaction is very common and is the main reaction when burning natural gas for fuel.

There are two types of methane combustion:

• Complete Combustion: Occurs when there is plenty of oxygen. Methane burns to produce carbon dioxide and water, releasing a lot of energy. The balanced chemical equation is:
  
  \(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\)
  
• Incomplete Combustion: Takes place when oxygen is limited. It produces carbon monoxide (a poisonous gas) and water. It's less efficient because it releases less energy compared to complete combustion. The balanced equation is:
  
  \(\text{CH}_4 + \frac{3}{2}\text{O}_2 \rightarrow \text{CO} + 2\text{H}_2\text{O}\)
  

Understanding these processes is crucial for ensuring the safe and efficient use of methane as an energy source.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes how gases behave under certain conditions. It relates four essential properties of gases: pressure, volume, temperature, and amount (in moles). The formula is:

• \(PV = nRT\)

Here, \(P\) is pressure, \(V\) is volume, \(n\) is the amount of gas in moles, \(R\) is the ideal gas constant, and \(T\) is temperature measured in Kelvin.

When working with gases, such as methane in combustion reactions, knowing the Ideal Gas Law helps calculate how much air or oxygen is necessary for combustion. In our case, we used this law to determine the oxygen needed to completely combust 1.0 L of methane. We found how many moles of gas are in that volume and used stoichiometry to find the required oxygen amount.

The Ideal Gas Law helps us calculate volume changes and predict gas behavior under different conditions, making it an invaluable tool in chemistry.