Let's break this problem into a step-by-step solution. We'll address each sub-question: --- ## **Given:** - **Gas composition (by volume):** - CO₂ = 6.5% - CH₄ = 49.5% - CO = 16.% - H₂ = 25.% - N₂ = 2.5% - O₂ = .5% - **Excess air:** 25% - **Standard conditions:** 21°C, 101.325 kPa --- ## **Step 1: Write Combustion Reactions** ### **CH₄:** \[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \] ### **CO:** \[ \text{CO} + .5\text{O}_2 \rightarrow \text{CO}_2 \] ### **H₂:** \[ \text{H}_2 + .5\text{O}_2 \rightarrow \text{H}_2\text{O} \] --- ## **Step 2: Assume a Basis** Assume **100 moles** of fuel gas (so % = moles): - CO₂: 6.5 mol - CH₄: 49.5 mol - CO: 16. mol - H₂: 25. mol - N₂: 2.5 mol - O₂: .5 mol --- ## **Step 3: Calculate Theoretical Oxygen Requirement** ### **For CH₄:** \[ 49.5 \text{ mol} \times 2 = 99 \text{ mol O}_2 \] ### **For CO:** \[ 16. \text{ mol} \times .5 = 8. \text{ mol O}_2 \] ### **For H₂:** \[ 25. \text{ mol} \times .5 = 12.5 \text{ mol O}_2 \] **Total theoretical O₂ needed:** \[ 99 + 8 + 12.5 = 119.5 \text{ mol} \] Subtract O₂ present in the fuel (.5 mol): \[ 119.5 - .5 = 119. \text{ mol O}_2 \text{ (net required)} \] --- ## **Step 4: Calculate Air Required** - **O₂ in air** = 21% by volume - **Air needed (theoretical, for 100 mol fuel):** \[ \text{Air} = \frac{119.}{.21} = 566.67 \text{ mol air} \] - **With 25% excess air:** \[ \text{Air supplied} = 566.67 \times 1.25 = 708.3 \text{ mol air} \] --- ## **Step 5: Air to Fuel Ratio (by Volume)** - Fuel gas: 100 mol - Air supplied: 708.3 mol **Air-to-fuel ratio (by volume):** \[ \boxed{7.08:1} \] --- ## **Step 6: Volume of Flue Gas from 10 Moles Fuel** ### **Calculate Flue Gas Components for 10 mol Fuel** Scale everything by .1 (since basis above is 100 mol): - Air supplied: 70.83 mol (for 10 mol fuel) #### **Dry Products:** - CO₂ from CH₄: 49.5 × .1 = 4.95 mol - CO₂ from CO: 16. × .1 = 1.6 mol - CO₂ in fuel: 6.5 × .1 = .65 mol - **Total CO₂:** \(4.95 + 1.6 + .65 = 7.2\) mol - N₂ from air: Air supplied × .79 = 70.83 × .79 = 55.96 mol - N₂ in fuel: 2.5 × .1 = .25 mol - **Total N₂:** \(55.96 + .25 = 56.21\) mol #### **Excess O₂:** - O₂ supplied: 70.83 × .21 = 14.87 mol - O₂ consumed: 11.9 mol (see above, for 10 mol fuel) - **Excess O₂:** 14.87 - 11.9 = 2.97 mol #### **H₂O:** - From CH₄: 2 × 4.95 = 9.90 mol - From H₂: 2 × 2.5 = 5. mol - **Total H₂O:** 9.90 + 5. = 14.90 mol ### **Total Flue Gas (wet):** \[ \text{CO}_2: 7.2\ \text{mol} \\ \text{N}_2: 56.21\ \text{mol} \\ \text{O}_2: 2.97\ \text{mol} \\ \text{H}_2\text{O}: 14.90\ \text{mol} \\ \text{Total:} = 7.2 + 56.21 + 2.97 + 14.90 = \boxed{81.28\ \text{mol}} \] --- ## **Step 7: Volume of Air (at 21°C and 101.325 kPa) for 10 mol Fuel** - Air supplied for 10 mol fuel = 70.83 mol - Use ideal gas law: \( V = \frac{nRT}{P} \) - \(R = .08206\ \text{L·atm/(mol·K)}\) - \(T = 21^\circ C = 294.15\ K\) - \(P = 101.325\ \text{kPa} = 1\ \text{atm}\) \[ V = nRT/P = 70.83 \times .08206 \times 294.15 / 1 = 1,710.38\ \text{L} = 1.71\ \text{m}^3 \] --- ## **Step 8: Partial Pressure of H₂O (at 200°C, 101.325 kPa)** - Total moles (from above): 81.28 mol - H₂O moles: 14.90 mol \[ P_{\text{H}_2\text{O}} = \frac{14.90}{81.28} \times 101.325\ \text{kPa} = 18.57\ \text{kPa} \] --- ## **Step 9: Volume of Exhaust Gas at 200°C, 101.325 kPa per 1 m³ Fuel at 20°C, 101.325 kPa** ### **Calculate moles of fuel per 1 m³ at 20°C, 101.325 kPa** - \( T = 20^\circ C = 293.15 K \) - \( P = 101.325 kPa \) - \( V = 1 m^3 = 100 L \) \[ n_{\text{fuel}} = \frac{PV}{RT} = \frac{101.325 \times 100}{8.314 \times 293.15} = 41.55\ \text{mol} \] ### **Moles of exhaust gas:** \[ 81.28\ \text{mol (flue) per 10 mol fuel} \implies 8.128\ \text{mol per 1 mol fuel} \] \[ \text{Total moles} = 41.55 \times 8.128 = 337.76\ \text{mol} \] ### **Volume at 200°C (473.15 K), 101.325 kPa** \[ V = \frac{nRT}{P} = \frac{337.76 \times 8.314 \times 473.15}{101.325} = 13,106.7\ \text{L} = 13.1\ \text{m}^3 \] --- ## **Summary of Answers** ### **a) Air to fuel ratio by volume:** \[ \boxed{7.08:1} \] ### **b) Volume of flue gas formed from 10 mol of fuel:** \[ \boxed{81.28\ \text{mol (wet)}} \] ### **c) Volume of air (at 21°C, 101.325 kPa) for 10 mol of fuel:** \[ \boxed{1.71\ \text{m}^3} \] ### **d) Partial pressure of H₂O in exhaust gas at 200°C, 101.325 kPa:** \[ \boxed{18.57\ \text{kPa}} \] ### **e) Volume of exhaust gas at 200°C, 101.325 kPa per 1 m³ fuel at 20°C, 101.325 kPa:** \[ \boxed{13.1\ \text{m}^3} \] --- If you want to see the calculations for a specific part in more depth, let me know!