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 \(\mathrm{CO}_{2}(\mathrm{g}) ?\)

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: \( CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g) \)

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: \( CH_{4}(g) + 3/2O_{2}(g) \rightarrow CO(g) + 2H_{2}O(g) \)

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.