Let's solve the problem step by step. --- ## 1. **Balanced Equation for Complete Combustion of NH₃** The complete combustion of ammonia (NH₃) with oxygen (O₂) produces nitrogen (N₂) and water (H₂O): \[ 4\text{NH}_3 + 3\text{O}_2 \rightarrow 2\text{N}_2 + 6\text{H}_2\text{O} \] --- ## 2. **Stoichiometric Oxygen and Air Requirement** ### a. **Oxygen Required (Stoichiometric)** From the balanced equation: - 4 moles NH₃ require 3 moles O₂ - 1 mole NH₃ requires \( \frac{3}{4} = .75 \) moles O₂ ### b. **Air Required** Air is ~21% O₂ by mole, so: \[ \text{Moles air required per mole NH}_3 = \frac{.75}{.21} = 3.57\ \text{moles air} \] --- ## 3. **Excess Air Calculation (200% Excess)** - **200% excess air** means total air supplied is **3 times** the stoichiometric amount: \[ \text{Total air supplied} = 3 \times 3.57 = 10.71\ \text{moles air per mole NH}_3 \] --- ## 4. **Product Gas Composition (per 1 mole NH₃)** Let's base the calculation on **1 mole NH₃**: ### a. **Reactants:** - NH₃: 1 mole - O₂ supplied: \( 10.71 \times .21 = 2.25 \) moles - N₂ supplied: \( 10.71 \times .79 = 8.46 \) moles ### b. **O₂ Consumed:** - .75 moles O₂ consumed (stoichiometrically) - Excess O₂ remaining: \( 2.25 - .75 = 1.50 \) moles ### c. **N₂ in Product:** - N₂ from air: 8.46 moles - N₂ from reaction: \( \frac{2}{4} = .5 \) moles per 1 mole NH₃ ### d. **H₂O Produced:** - \( \frac{6}{4} = 1.5 \) moles per 1 mole NH₃ #### **Summary Table:** | Component | Amount (mol) | |-----------|--------------| | N₂ (from air) | 8.46 | | N₂ (from reaction) | .5 | | O₂ (excess) | 1.50 | | H₂O | 1.5 | | **Total** | **11.96** | --- ## 5. **Mole Fraction of H₂O** \[ \text{Mole fraction of } \text{H}_2\text{O} = \frac{1.5}{11.96} = .125 \] --- ## 6. **Partial Pressure of H₂O** Let total pressure \( P_{total} = 1 \) atm (unless stated otherwise): \[ P_{\text{H}_2\text{O}} = y_{\text{H}_2\text{O}} \cdot P_{total} = .125 \times 1 = .125\ \text{atm} \] --- ## 7. **Dew Point Calculation** - Dew point: The temperature at which water vapor starts to condense (saturated vapor pressure equals partial pressure). - Use steam tables or Antoine equation for \( P_{\text{H}_2\text{O}} = .125 \) atm. At \( P_{\text{H}_2\text{O}} = .125 \) atm: - \( .125 \) atm \( = 95 \) mmHg From steam tables, the temperature at which vapor pressure of water is \( 95 \) mmHg is approximately **49°C**. --- ## **Summary Table** | Quantity | Value | |---------------------|--------------| | Balanced Equation | \(4\text{NH}_3 + 3\text{O}_2 \rightarrow 2\text{N}_2 + 6\text{H}_2\text{O}\) | | H₂O Mole Fraction | .125 | | H₂O Partial Pressure| .125 atm | | Dew Point | 49°C | --- ### **Final Answers** - **Balanced Equation:** \(4\text{NH}_3 + 3\text{O}_2 \rightarrow 2\text{N}_2 + 6\text{H}_2\text{O}\) - **H₂O Mole Fraction:** .125 - **H₂O Partial Pressure:** .125 atm - **Dew Point:** 49°C If you need calculations for a different basis (e.g., 100 mol NH₃), just multiply accordingly!