Question

Natural gas consists primarily of methane, \(\mathrm{CH}_{4}(\mathrm{~g})\). (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 \(\mathrm{CO}(\mathrm{g})\) as the only carbon-containing product. (c) At \(25^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\) 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 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 equation for the incomplete combustion of methane is: \(CH_{4}(g) + \frac{3}{2}O_{2}(g) \rightarrow CO(g) + 2H_{2}O(g)\). (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 3.10 L.

Step-by-step solution

(a) Complete Combustion of Methane equation

A complete combustion of methane (CH4) occurs when it reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The unbalanced equation for a complete combustion is: CH4(g) + O2(g) → CO2(g) + H2O(g) To balance the equation, we need to make sure the number of each type of atom on both sides of the equation is the same. We can do this by adding stoichiometric coefficients to balance the number of atoms: \(CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)\) This equation is now balanced.

(b) Incomplete Combustion of Methane equation

An incomplete combustion of methane occurs when it reacts with oxygen (O2) to produce carbon monoxide (CO) and water (H2O). The unbalanced equation for an incomplete combustion is: CH4(g) + O2(g) → CO(g) + H2O(g) To balance the equation, we need to make sure the number of each type of atom on both sides of the equation is the same. We can do this by adding stoichiometric coefficients to balance the atoms: \(CH_{4}(g) + \frac{3}{2}O_{2}(g) \rightarrow CO(g) + 2H_{2}O(g)\) This equation is also balanced.

(c) Volume of dry air needed for complete combustion

To find the minimum volume of dry air required for the complete combustion of 1.0 L of CH4(g) at 25°C and 1.0 atm pressure, we first need to determine the amount of O2 needed. In the balanced equation for the complete combustion, we found that 1 mole of CH4 requires 2 moles of O2 for complete combustion. First, we need to find the number of moles of CH4 in 1.0 L. We can use the Ideal Gas Law equation: \(PV = nRT\) Where P is the pressure (1.0 atm), V is the volume (1.0 L), n is the number of moles, R is the ideal gas constant (0.0821 L atm/(mol K)), and T is the temperature (25°C = 298 K). We can isolate n to find the number of moles: \(n=\frac{PV}{RT}=\frac{(1.0\,atm)(1.0\,L)}{(0.0821\,L\,atm/(mol\,K))(298\,K)}\) \(n = 0.0409\,mol\) Now, we know that 2 moles of O₂ are needed for every mole of CH₄, so we can find the moles of O₂ required: \(0.0409\,mol\,CH_{4} \times \frac{2\,mol\,O_{2}}{1\,mol\,CH_{4}}=0.0818\,mol\,O_{2}\) We also know that dry air contains approximately 21% oxygen by volume. So we can find the total volume of air needed by dividing the volume of O₂ by 0.21: \(V_{air} = \frac{0.0818\,mol\,O_{2}}{0.21}\) Now we can use the Ideal Gas Law again to find the volume of dry air needed: \(V_{air} = \frac{n_{O_2}RT}{P} = \frac{(0.0818\,mol)(0.0821\,L\,atm/(mol\,K))(298\,K)}{1.0\,atm}\) After calculating, we get: \(V_{air} = 3.10\,L\) 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 3.10 L.

Key concepts

Chemical Equations

Understanding the essentials of chemical equations is fundamental to grasping the concept of chemical reactions like the combustion of methane. A chemical equation represents the change from reactants to products, showing the substances involved and how they are transformed. It's crucial to balance chemical equations to adhere to the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.

For example, the complete combustion of methane is demonstrated by the balanced equation \(CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)\). In this equation, the number of carbon, hydrogen, and oxygen atoms is the same on both sides, satisfying the balancing requirement. Chemical equations also provide insight into reaction stoichiometry, which is the study of the relative quantities of reactants and products in a chemical reaction.

Stoichiometry

Stoichiometry is the quantitative aspect of chemical equations. It helps us determine the amounts of substances involved in a reaction. The balanced chemical equations serve as a map for stoichiometry, allowing us to calculate necessary reactants or products.

In the exercise, the stoichiometric relationship gives a clear indication that one mole of methane reacts with two moles of oxygen to give one mole of carbon dioxide. This ratio is crucial when scaling up reactions and ensuring that reactants are used efficiently. Understanding stoichiometry involves comprehending that reactions occur in fixed proportions, a concept which is essential when dealing with chemical reactions in real-world applications.

Ideal Gas Law

The Ideal Gas Law is a critical concept in understanding the relationship between the pressure, volume, temperature, and number of moles of a gas. The equation \(PV = nRT\) encapsulates this law, where \(P\) stands for pressure, \(V\) for volume, \(n\) for the number of moles of the gas, \(R\) for the ideal gas constant, and \(T\) for temperature in Kelvin.

Through this law, we can deduce that for each litre of methane combusted at standard conditions, we need a fixed volume of air (which contains the oxygen required for the reaction). By calculating the moles of methane using the Ideal Gas Law, one can then work out the amount of dry air necessary for combustion. For example, to combust 1.0 L of methane, we need 3.10 L of dry air, considering the proportion of oxygen in air and the stoichiometry of the reaction. The Ideal Gas Law thus serves as a bridge linking the theoretical volumes of gases to practical conditions that guide experimental and industrial processes.