Let's break down the problem shown in the image: --- ### Problem Summary - 1 kmole of a hydrocarbon fuel \( C_nH_{2n+2} \) (with \( n = 8 \), so **C₈H₁₈**) is burned with an unknown amount of air. - After combustion, 3 kmoles of O₂ are found in the products (excess oxygen). - Find the **quantity of air involved** and the **air-fuel ratio (AFR)**. --- ### Step 1: Write the Combustion Equation For **n = 8**: \[ \mathrm{C_8H_{18}} + a(\mathrm{O_2} + 3.76\ \mathrm{N_2}) \rightarrow 8\ \mathrm{CO_2} + 9\ \mathrm{H_2O} + x\ \mathrm{O_2} + y\ \mathrm{N_2} \] --- ### Step 2: Stoichiometric (Theoretical) Oxygen Required \[ \mathrm{C_8H_{18}} + \left( \frac{8 + \frac{18}{4}}{1} \right) \mathrm{O_2} \rightarrow \text{products} \] \[ \text{O}_2\ \text{needed} = 8 + \frac{18}{4} = 8 + 4.5 = 12.5\ \text{kmol O}_2 \] --- ### Step 3: Actual Oxygen Supplied There is 3 kmol O₂ left after combustion, so: \[ \text{O}_2\ \text{supplied} = \text{O}_2\ \text{needed} + \text{excess O}_2 = 12.5 + 3 = 15.5\ \text{kmol O}_2 \] --- ### Step 4: Air Supplied Air contains 21% O₂ by volume (by mole): \[ \text{Air supplied} = \frac{15.5}{.21} \approx 73.81\ \text{kmol air} \] Or, more commonly, using mass ratio: - 1 kmol air contains 1 kmol O₂ + 3.76 kmol N₂ = 4.76 kmol (total "air" per O₂ mol) \[ \text{Air supplied} = 15.5 \times 4.76 = 73.78\ \text{kmol air} \] --- ### Step 5: Excess Air \[ \text{Excess air} = \frac{\text{Actual O}_2\ \text{supplied} - \text{Theoretical O}_2}{\text{Theoretical O}_2} \times 100\% \] \[ = \frac{15.5 - 12.5}{12.5} \times 100 = \frac{3}{12.5} \times 100 = 24\% \] --- ### Step 6: Air-Fuel Ratio (AFR) - 1 kmol C₈H₁₈: molecular mass = \( 8\times12 + 18\times1 = 114 \) kg - 1 kmol air = 28.96 kg - Total air supplied (kmol): \( 15.5 \times 4.76 = 73.78 \) kmol - Mass of air: \( 73.78 \times 28.96 = 2138.5 \) kg \[ \text{AFR} = \frac{\text{Mass of air}}{\text{Mass of fuel}} = \frac{2138.5}{114} \approx 18.76 \] --- ### Step 7: Find the Correct Option Let's check the options: 1. **24% of excess air and AFR = 14.75** 2. **124% of excess air and AFR = 18.76** 3. **124% of theoretical air and AFR = 18.76** - **24% excess air and AFR = 18.76** is the correct calculation. - **124% of theoretical air** means supplying 1.24 times the theoretical air, i.e., 24% excess air. So, this is also correct: **124% of theoretical air and AFR = 18.76** **Correct Option:** > **"124% of theoretical air and AFR = 18.76"** --- ## Final Answer **The correct option is:** > **124% of theoretical air and AFR = 18.76**