# Benzene-Toluene Distillation Problem: Step-by-Step Solution Given: - **Feed rate (\(F\))**: 120ol/h - **Feed benzene mole fraction (\(x_F\))**: .45 - **Distillate benzene mole fraction (\(x_D\))**: .92 - **Bottoms benzene mole fraction (\(x_B\))**: .08 - **Assumptions**: Steady state, negligible heat loss, constant molar overflow, ideal behavior, no reaction. --- ## (a) Molar Flow Rate of Distillate (\(D\)) ### Step 1: Write Material Balances #### Total Material Balance: \[ F = D + B \] #### Benzene Material Balance: \[ F x_F = D x_D + B x_B \] ### Step 2: Substitute Known Values - \(F = 120\) kmol/h - \(x_F = .45\) - \(x_D = .92\) - \(x_B = .08\) From the total balance: \[ B = F - D \] Substitute \(B\) into the benzene balance: \[ F x_F = D x_D + (F - D) x_B \] \[ F x_F = D x_D + F x_B - D x_B \] \[ F x_F - F x_B = D(x_D x_B) \] \[ D = \frac{F(x_F - x_B)}{x_D - x} \] ### Step 3: Calculate \(D\) \[ D = \frac{120 \times (.45 - .08)}{.92 - .08} \] \[ D = \frac{120 \times .37}{.84} \] \[ D = \frac{44.4}{.84} \] \[ D = 52.857~\text{kmol/h} \] **Final Answer (a):** \[ \boxed{52.86~\text{kmol/h}} \] --- ## (b) Molar Flow Rate of Bottoms (\(B\)) \[ B = F - D = 120 - 52.86 = 67.14~\text{kmol/h} \] **Final Answer (b):** \[ \boxed{67.14~\text{kmol/h}} \] --- ## (c) Benzene Recovery Percentage in Distillate \[ \text{Benzene in distillate} = D \times x_D = 52.86 \times .92 = 48.63~\text{kmol/h} \] \[ \text{Benzene in feed} = F \times x_F = 120 \times .45 = 54~\text{kmol/h} \] \[ \text{Recovery (\%)} = \frac{48.63}{54} \times 100 = 90.06\% \] **Final Answer (c):** \[ \boxed{90.1\%} \] --- ## (d) Molar Flow Rate of Toluene in Distillate - Mole fraction of toluene in distillate: \(1 - x_D = 1 - .92 = .08\) \[ \text{Toluene in distillate} = D \times (1 - x_D) = 52.86 \times .08 = 4.23~\text{kmol/h} \] **Final Answer (d):** \[ \boxed{4.23~\text{kmol/h}} \] --- ## (e) Mole Fraction of Toluene in Bottoms - Mole fraction of benzene in bottoms: \(x_B = .08\) - Thus, mole fraction of toluene in bottoms: \(1 - x_B = 1 - .08 = .92\) **Final Answer (e):** \[ \boxed{.92} \] --- ## (f) New Distillate Flow Rate if Feed Increases by 20% ### Step 1: Calculate New Feed Rate \[ F_{\text{new}} = 120 \times 1.20 = 144~\text{kmol/h} \] ### Step 2: Use the Same Compositions \[ D_{\text{new}} = \frac{F_{\text{new}}(x_F - x_B)}{x_D - x_B} \] \[ D_{\text{new}} = \frac{144 \times .37}{.84} \] \[ D_{\text{new}} = \frac{53.28}{.84} \] \[ D_{\text{new}} = 63.43~\text{kmol/h} \] **Final Answer (f):** \[ \boxed{63.43~\text{kmol/h}} \] --- ## Summary Table | Part | Quantity | Value | Unit | |------|--------------------------------------------|---------------|--------------| | (a) | Distillate flow rate (\(D\)) | 52.86 | kmol/h | | (b) | Bottoms flow rate (\(B\)) | 67.14 | kmol/h | | (c) | Benzene recovery in distillate | 90.1 | % | | (d) | Toluene in distillate | 4.23 | kmol/h | | (e) | Toluene fraction in bottoms | .92 | mole fraction| | (f) | New distillate flow (20% more feed) | 63.43 | kmol/h | --- ### Note - All calculations assume no losses and ideal, constant-molar overflow. - Answers are rounded to two decimal places for clarity.