# Complete Solution: Combustion of NH₃ with 200% Excess Air Let's solve **step-by-step**: --- ## 1. **Balanced Equation for Complete Combustion of NH₃** Ammonia (NH₃) combusts as: \[ 4\text{NH}_3 + 3\text{O}_2 \rightarrow 2\text{N}_2 + 6\text{H}_2\text{O} \] --- ## 2. **Stoichiometric Oxygen and Air Required** **For 1 mol NH₃:** - \(4\text{NH}_3\) needs \(3\text{O}_2\) - \(\Rightarrow 1\text{NH}_3\) needs \(\frac{3}{4}\text{O}_2 = .75\text{O}_2\) **Air contains 21% O₂ by mole:** - 1 mol O₂ from \(\frac{1}{.21} = 4.76\) mol air **Stoichiometric air per 1 mol NH₃**: \[ .75\text{ mol O}_2 \times 4.76 = 3.57 \text{ mol air} \] --- ## 3. **200% Excess Air** "200% excess" means **total air supplied = 3 × stoichiometric air**: \[ \text{Total air supplied} = 3 \times 3.57 = 10.71 \text{ mol air per 1 mol NH}_3 \] --- ## 4. **Determine Product Gas Composition (per 1 mol NH₃)** ### a. **Reactants** - 1 mol NH₃ - 10.71 mol air (contains O₂ and N₂) ### b. **Oxygen in Supplied Air** - \(10.71 \times .21 = 2.25\) mol O₂ ### c. **Nitrogen in Supplied Air** - \(10.71 \times .79 = 8.46\) mol N₂ ### d. **Combustion Reaction for 1 mol NH₃** \[ \text{NH}_3 + .75\text{O}_2 \rightarrow .5\text{N}_2 + 1.5\text{H}_2\text{O} \] ### e. **O₂ Remaining (Excess O₂)** - Supplied: 2.25 mol O₂ - Used: .75 mol O₂ - **Excess O₂ left in product**: \(2.25 - .75 = 1.50\) mol O₂ ### f. **N₂ in Products** - From air: 8.46 mol N₂ - From reaction: .5 mol N₂ - **Total N₂**: \(8.46 + .5 = 8.96\) mol N₂ ### g. **H₂O Produced** - From reaction: 1.5 mol H₂O ### h. **Product Gas Totals (per 1 mol NH₃ combusted)** | Species | Amount (mol) | |---------|--------------| | N₂ | 8.96 | | O₂ | 1.50 | | H₂O | 1.5 | | **Total** | **11.96** | --- ## 5. **H₂O Molar Fraction** \[ y_{\text{H}_2\text{O}} = \frac{1.5}{11.96} = .125 \] --- ## 6. **H₂O Partial Pressure** Assume total pressure = 1 atm (unless stated otherwise): \[ P_{\text{H}_2\text{O}} = y_{\text{H}_2\text{O}} \times P_{\text{total}} = .125 \times 1 = .125~\text{atm} \] --- ## 7. **Dew Point Calculation** - Dew point: temperature at which partial pressure of H₂O equals saturated vapor pressure. Look up the temperature where vapor pressure of water = .125 atm. From steam tables: - At **49°C (120°F)**, vapor pressure of water ≈ .123 atm - At **50°C (122°F)**, vapor pressure ≈ .128 atm So, **dew point ≈ 50°C**. --- ## **Final Summary Table** | Quantity | Value | |-----------------------|--------------| | **Balanced Equation** | \(4 \text{NH}_3 + 3 \text{O}_2 \rightarrow 2 \text{N}_2 + 6 \text{H}_2\text{O}\) | | **Air Supplied** | 10.71 mol per 1 mol NH₃ | | **H₂O Molar Fraction**| .125 | | **H₂O Partial Pressure**| .125 atm | | **Dew Point** | ~50°C | --- **If you need the answer for a different amount of NH₃ or at a different pressure, let me know!**