## Step 1: Write the Full Chemical Equation the Combustion (Part a) **Combustion of Ethylene (C₂H₄) with Stoichiometric Air:** The balanced chemical equation for the complete combustion of ethylene with oxygen is: \[ \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \] Since air is used (which contains 21% O₂ and 79% N₂ by mole), the equation with air becomes: \[ \text{C}_2\text{H}_4 + 3(\text{O}_2 + 3.76\text{N}_2) \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} + 3 \times 3.76\text{N}_2 \] Where 3.76 comes from the molar ratio of N₂ to O₂ in air (79/21 ≈ 3.76). **Explanation:** This step shows how to set up the correct stoichiometric equation for ethylene burning with air, accounting for both oxygen and nitrogen. --- ## Step 2: For 1 kg Consumption of Fuel, Determine the Mass of CO₂ Produced (Part b) **Step 2a: Find Number of Moles of C₂H₄ in 1 kg** Molar mass of C₂H₄ = (2 × 12) + (4 × 1) = 28 kg/kmol Number of moles in 1 kg: \[ \text{Moles of C}_2\text{H}_4 = \frac{1 \text{ kg}}{28 \text{ kg/kmol}} = .03571 \text{ kmol} \] **Step 2b: From the Equation, 1 mol C₂H₄ produces 2 mol CO₂** So, \[ \text{Moles of CO}_2 = 2 \times \text{Moles of C}_2\text{H}_4 = 2 \times .03571 = .07143 \text{ kmol} \] Molar mass of CO₂ = 44 kg/kmol \[ \text{Mass of CO}_2 = .07143 \text{ kmol} \times 44 \text{ kg/kmol} = 3.143 \text{ kg} \] **Explanation:** We first calculated the amount of ethylene in 1 kg, then used the stoichiometry to find the amount of CO₂ produced, and finally converted it to mass using the molar mass. --- ## Final Answers **(a) Full Chemical Equation:** \[ \text{C}_2\text{H}_4 + 3(\text{O}_2 + 3.76\text{N}_2) \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} + 11.28\text{N}_2 \] **(b) Mass of CO₂ Produced per 1 kg Ethylene:** \[ \boxed{3.14\ \text{kg}} \]

## Step 1: Write the Full Chemical Equation for the Combustion (Part a) **Combustion of Ethylene (C₂H₄) with Stoichiometric Air:** The balanced chemical equation for the complete combustion of ethylene with oxygen is: \[ \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \] When air is used (which contains approximately 21% O₂ and 79% N₂), the equation becomes: \[ \text{C}_2\text{H}_4 + 3(\text{O}_2 + 3.76\text{N}_2) \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} + 11.28\text{N}_2 \] Where the 3.76 factor is derived from the molar ratio of N₂ to O₂ in air (approximately 79/21). **Explanation:** This step illustrates the complete stoichiometric equation for the combustion of ethylene with air, accounting for both oxygen and nitrogen components. --- ## Step 2: Determine the Mass of CO₂ Produced for 1 kg Consumption of Fuel (Part b) **Step 2a: Calculate the Number of Moles of C₂H₄ in 1 kg** The molar mass of C₂H₄ is calculated as follows: \[ \text{Molar mass of C}_2\text{H}_4 = (2 \times 12) + (4 \times 1) = 28 \text{ kg/kmol} \] The number of moles in 1 kg of ethylene is: \[ \text{Moles of C}_2\text{H}_4 = \frac{1 \text{ kg}}{28 \text{ kg/kmol}} = 0.03571 \text{ kmol} \] **Step 2b: Calculate the Moles of CO₂ Produced** From the balanced equation, 1 mole of C₂H₄ produces 2 moles of CO₂: \[ \text{Moles of CO}_2 = 2 \times \text{Moles of C}_2\text{H}_4 = 2 \times 0.03571 = 0.07143 \text{ kmol} \] Next, we calculate the mass of CO₂ produced using its molar mass (44 kg/kmol): \[ \text{Mass of CO}_2 = 0.07143 \text{ kmol} \times 44 \text{ kg/kmol} = 3.143 \text{ kg} \] **Explanation:** We first determined the amount of ethylene present in 1 kg, then applied stoichiometric ratios to find the amount of CO₂ produced, converting it to mass using the molar mass of CO₂. --- ## Final Answers **(a) Full Chemical Equation:** \[ \text{C}_2\text{H}_4 + 3(\text{O}_2 + 3.76\text{N}_2) \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} + 11.28\text{N}_2 \] **(b) Mass of CO₂ Produced per 1 kg Ethylene:** \[ \boxed{3.14\ \text{kg}} \]