Let's solve the problem step by step: --- ## Given Reaction: \[ \mathrm{C_4H_{10}} + 8(\mathrm{O_2} + 3.76\mathrm{N_2}) \rightarrow 4\mathrm{CO_2} + 5\mathrm{H_2O} + 1.5\mathrm{O_2} + 8(3.76)\mathrm{N_2} \] ### (a) **Fuel-to-air ratio,** \( f \) #### **Step 1: Find moles of fuel and air** - 1 mole of butane (\(\mathrm{C_4H_{10}}\)) - Air: \(8(\mathrm{O_2} + 3.76\mathrm{N_2})\) #### **Step 2: Calculate moles of air** - 1 mole of \(\mathrm{O_2}\) is accompanied by 3.76 moles of \(\mathrm{N_2}\) in air. - Moles of air \(= 8 \times (1 + 3.76) = 8 \times 4.76 = 38.08\) moles of air #### **Step 3: Find the mass of fuel and air** **Molar mass of butane (\(\mathrm{C_4H_{10}}\)):** \[ \mathrm{C_4H_{10}}: (4 \times 12.01) + (10 \times 1.008) = 48.04 + 10.08 = 58.12~\mathrm{g/mol} \] **Molar mass of air:** \[ \text{Average molar mass of dry air} \approx 28.97~\mathrm{g/mol} \] #### **Step 4: Calculate fuel-to-air ratio (\( f \))** \[ f = \frac{\text{mass of fuel}}{\text{mass of air}} \] \[ \text{Mass of fuel} = 1~\mathrm{mol} \times 58.12~\mathrm{g/mol} = 58.12~\mathrm{g} \] \[ \text{Mass of air} = 38.08~\mathrm{mol} \times 28.97~\mathrm{g/mol} = 1103~\mathrm{g} \] \[ f = \frac{58.12}{1103} = .0527 \] --- ### (b) **Equivalence ratio, \( \phi \)** \[ \phi = \frac{(\text{fuel/air})_\text{actual}}{(\text{fuel/air})_\text{stoichiometric}} \] #### **Step 1: Find stoichiometric fuel-to-air ratio** Stoichiometric combustion of butane: \[ \mathrm{C_4H_{10}} + x (\mathrm{O_2} + 3.76 \mathrm{N_2}) \rightarrow \] Complete combustion: \[ \mathrm{C_4H_{10}} + \frac{13}{2}\mathrm{O_2} + \frac{13}{2} \cdot 3.76 \mathrm{N_2} \rightarrow 4\mathrm{CO_2} + 5\mathrm{H_2O} + (\text{N}_2) \] So, stoichiometric air required: \[ \frac{13}{2}(\mathrm{O_2} + 3.76\mathrm{N_2}) = 6.5 \times 4.76 = 30.94~\text{mol air} \] \[ \text{Mass of air (stoich)} = 30.94 \times 28.97 = 897~\mathrm{g} \] \[ f_{stoich} = \frac{58.12}{897} = .0648 \] #### **Step 2: Calculate equivalence ratio \( \phi \)** \[ \phi = \frac{f_{actual}}{f_{stoich}} = \frac{.0527}{.0648} = .813 \] --- ## **Final Answers** **(a) Fuel-to-air ratio, \( f \):** \[ \boxed{.0527} \] **(b) Equivalence ratio, \( \phi \):** \[ \boxed{.813} \] --- Let me know if you'd like any clarifications!