Sure! Let’s break down the problem step by step and explain **how to find the mole fraction of water vapor in the products** (Step 3), given the information in the image. --- ## Problem Recap A natural gas mixture (by mole): - 78% CH₄, 13% C₂H₆, 6% C₃H₈, 1.7% C₄H₁₀, 1.3% N₂ **Burns completely** with **30% excess air**. - Fuel molar flow rate: 1.5 kmol/h - Air composition: 21% O₂, 79% N₂ (by mole) - Air molar flow rate: 22.5 kmol/h (from Step 1) --- # Step 1: **Write the Combustion Reactions** Each hydrocarbon combusts as: - CH₄ + 2O₂ → CO₂ + 2H₂O - C₂H₆ + 3.5O₂ → 2CO₂ + 3H₂O - C₃H₈ + 5O₂ → 3CO₂ + 4H₂O - C₄H₁₀ + 6.5O₂ → 4CO₂ + 5H₂O --- # Step 2: **Calculate Moles of Each Fuel Component** Given **total fuel flow = 1.5 kmol/h**: | Component | Mole Fraction | Moles/hr | | ---- | ---- | ---- | | CH₄ | .78 | 1.5 × .78 = **1.17 kmol/h** | | C₂H₆ | .13 | 1.5 × .13 = **.195 kmol/h** | | C₃H₈ | .06 | 1.5 × .06 = **.09 kmol/h** | | C₄H₁₀ | .017 | 1.5 × .017 = **.0255 kmol/h** | | N₂ | .013 | 1.5 × .013 = **.0195 kmol/h** | --- # Step 3: **Calculate Moles of Each Product (H₂O, CO₂, etc.)** ## **a. Find H₂O Produced** - **CH₄:** 1.17 kmol × 2 = **2.34 kmol H₂O** - **C₂H₆:** .195 kmol × 3 = **.585 kmol H₂O** - **C₃H₈:** .09 kmol × 4 = **.36 kmol H₂O** - **C₄H₁₀:** .0255 kmol × 5 = **.1275 kmol H₂O** **Total H₂O produced:** \[ 2.34 + .585 + .36 + .1275 = \boxed{3.4125\ \text{kmol H}_2\text{O}} \] --- ## **b. Find Other Product Moles** - **CO₂ produced:** - CH₄: 1.17 kmol × 1 = 1.17 kmol - C₂H₆: .195 kmol × 2 = .39 kmol - C₃H₈: .09 kmol × 3 = .27 kmol - C₄H₁₀: .0255 kmol × 4 = .102 kmol - **Total CO₂:** 1.17 + .39 + .27 + .102 = **1.932 kmol** - **N₂ in products:** 1. From fuel: .0195 kmol 2. From air: Air is 22.5 kmol, 79% is N₂ ⇒ 22.5 × .79 = 17.775 kmol 3. Total N₂: .0195 + 17.775 = **17.7945 kmol** - **O₂ in products:** 1. Find total O₂ supplied: Air is 21% O₂ ⇒ 22.5 × .21 = **4.725 kmol O₂ supplied** 2. Subtract O₂ consumed (see below) --- ## **c. Calculate Theoretical and Actual O₂ Required** ### **Theoretical O₂ demand:** - **CH₄:** 1.17 kmol × 2 = 2.34 kmol O₂ - **C₂H₆:** .195 kmol × 3.5 = .6825 kmol O₂ - **C₃H₈:** .09 kmol × 5 = .45 kmol O₂ - **C₄H₁₀:** .0255 kmol × 6.5 = .16575 kmol O₂ **Total O₂ required (stoichiometric):** 2.34 + .6825 + .45 + .16575 = **3.63825 kmol O₂** ### **With 30% Excess Air:** \[ \text{Actual O}_2 = 3.63825 \times 1.3 = 4.729725\ \text{kmol} \] (Small rounding difference with air's O₂, but matches the 4.725 kmol supplied.) ### **Excess O₂ in Products:** \[ \text{Excess O}_2 = \text{O}_2\ \text{supplied} - \text{O}_2\ \text{required} = 4.725 - 3.63825 = 1.08675\ \text{kmol} \] --- # Step 4: **Calculate Total Moles of Product** Add up all the products: - H₂O: 3.4125 kmol - CO₂: 1.932 kmol - N₂: 17.7945 kmol - O₂ (excess): 1.08675 kmol \[ \text{Total moles in products} = 3.4125 + 1.932 + 17.7945 + 1.08675 = \boxed{24.22575\ \text{kmol}} \] --- # Step 5: **Calculate Mole Fraction of Water Vapor** \[ x_{\text{H}_2\text{O}} = \frac{\text{kmol H}_2\text{O}}{\text{Total kmol products}} = \frac{3.4125}{24.22575} = \boxed{.141} \] --- ## **Final Answer** ### **The mole fraction of water vapor in the products is:** \[ \boxed{.141} \] --- ## **Summary Table** | Species | Mole in Products (kmol) | |---------|------------------------| | H₂O | 3.4125 | | CO₂ | 1.932 | | N₂ | 17.7945 | | O₂ | 1.08675 | | **Total** | 24.22575 | --- ## **Key Steps to Remember** 1. Write balanced combustion equations. 2. Calculate moles of each fuel component. 3. Determine moles of each product, including N₂ and O₂. 4. Add up all product moles. 5. Divide moles of H₂O by total product moles for mole fraction. --- If you need any step explained in more detail, let me know!