Question

One \(\mathrm{kmol}\) of methane \(\left(\mathrm{CH}_{4}\right)\) is burned with an unknown amount of air during a combustion process. If the combustion is complete and there are \(1 \mathrm{kmol}\) of free \(\mathrm{O}_{2}\) in the products, the air-fuel mass ratio is \((a) 34.6\) (b) 25.7 \((c) 17.2\) \((d) 14.3\) \((e) 11.9\)

Short answer

Answer: (a) 34.6

Step-by-step solution

Write the balanced chemical reaction

We have the complete combustion of methane (\(\mathrm{CH}_4\)) in the presence of oxygen (\(\mathrm{O}_2\)), producing carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)). The balanced chemical equation for this reaction is: $$\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}$$

Calculate the required oxygen for complete combustion of 1 kmol of methane

From the balanced chemical equation, 2 kmol of \(\mathrm{O}_2\) are required for the complete combustion of 1 kmol of \(\mathrm{CH}_4\). Thus, the required amount of \(\mathrm{O}_2\) is 2 kmol.

Calculate the total oxygen in air after combustion

Since we have 1 kmol of free \(\mathrm{O}_2\) in the products, the total amount of \(\mathrm{O}_2\) in air after combustion is the required amount of \(\mathrm{O}_2\) for complete combustion plus the free \(\mathrm{O}_2\). Therefore, Total \(\mathrm{O}_2\) in air = Required \(\mathrm{O}_2\) for complete combustion + Free \(\mathrm{O}_2\) Total \(\mathrm{O}_2\) in air = 2 kmol + 1 kmol = 3 kmol

Calculate the mass of air required for complete combustion

We know that air is composed of 21% \(\mathrm{O}_2\) and 79% \(\mathrm{N}_2\) by volume. To find the total mass of air, we need to determine the mass of both oxygen and nitrogen present in the air. First, we calculate the total mass of oxygen: Total mass of \(\mathrm{O}_2\) = Total \(\mathrm{O}_2\) in air × Molar mass of \(\mathrm{O}_2\) Total mass of \(\mathrm{O}_2\) = 3 kmol × 32 kg/kmol = 96 kg Next, we calculate the total mass of nitrogen in the air: Since the proportion of \(\mathrm{O}_2\) in air is 21%, we can determine the proportion of \(\mathrm{N}_2\) by: $$\frac{\mathrm{Mass\; of\; N_2}}{\mathrm{Mass\; of\; N_2} + 96 kg} = 0.79$$ Solving for the mass of \(\mathrm{N}_2\), we get: $$\mathrm{Mass\; of\; N}_2 = \frac{96 kg}{1-0.79} = 457.14 kg$$ Now, we can find the total mass of air: Total mass of air = Total mass of \(\mathrm{O}_2\) + Total mass of \(\mathrm{N}_2\) Total mass of air = 96 kg + 457.14 kg = 553.14 kg

Calculate the air-fuel mass ratio and determine the answer

The air-fuel mass ratio is the ratio of the mass of air required to the mass of fuel. In this case, the mass of fuel is the mass of 1 kmol of methane. We can determine the molar mass of methane as follows: Molar mass of \(\mathrm{CH}_4\) = 12 kg/kmol (carbon) + 4 × 1 kg/kmol (hydrogen) = 16 kg/kmol So, the mass of 1 kmol methane is 16 kg. Now, we calculate the air-fuel mass ratio: $$\mathrm{Air}\text{-}\mathrm{fuel}\;\mathrm{mass}\;\mathrm{ratio} = \frac{\mathrm{Total\; mass\; of\; air}}{\mathrm{Mass\; of}\; 1 \mathrm{kmol}\;\mathrm{of}\; \mathrm{CH}_4} = \frac{553.14\;\mathrm{kg}}{16\;\mathrm{kg}} = 34.57$$ Inspecting the given options, the air-fuel mass ratio closest to 34.57 is (a) 34.6. Therefore, the correct answer is (a) 34.6.