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Q3.7 For the combustion of methane presented in Example 5.4 (of the recommended text – Solen & Harb), the chemical reaction is CH4 +202 → CO2 + 2H2O Let's suppose that methane flows into a burner at 30 mol/s, while oxygen flows into the same burner at 75 mol/s. a) Which reactant is limiting? b) What is the percentage excess of the other reactant? C) If all of the methane is burned, and a single output stream leaves the burner, what is the mole fraction of CO2 in that output stream? [Adapted from Chapter 5, Question 13 of Solen & Harb] Q3.8 A chemical process is used to convert toluene (C7H8) and hydrogen to benzene (C6H6) and methane by the following reaction: C7H3 + H2 → C6H6 + CH4 Two streams enter the process. The first input stream is pure liquid toluene, which enters at a rate of 40 mol/s. The second input stream is a mixture of H2 (95 mole%) and CH4 (5 mole%). The flow rate of H2 in the second stream is equal to 200 mol/s. This process will include reaction and separation, and two streams will exit the process. The first output stream contains only liquid benzene (product) and toluene (unconverted reactant). The second output stream contains gaseous H2 and CH4. If the conversion of toluene in the process is 75%, determine the following: a) the molar flow rate and mass fraction of benzene in the liquid output stream b) the mole fraction and mass fraction of CH4 in the output gas stream? Show transcribed image text Q3.7 For the combustion of methane presented in Example 5.4 (of the recommended text – Solen & Harb), the chemical reaction is CH4 +202 → CO2 + 2H2O Let's suppose that methane flows into a burner at 30 mol/s, while oxygen flows into the same burner at 75 mol/s. a) Which reactant is limiting? b) What is the percentage excess of the other reactant? C) If all of the methane is burned, and a single output stream leaves the burner, what is the mole fraction of CO2 in that output stream? [Adapted from Chapter 5, Question 13 of Solen & Harb] Q3.8 A chemical process is used to convert toluene (C7H8) and hydrogen to benzene (C6H6) and methane by the following reaction: C7H3 + H2 → C6H6 + CH4 Two streams enter the process. The first input stream is pure liquid toluene, which enters at a rate of 40 mol/s. The second input stream is a mixture of H2 (95 mole%) and CH4 (5 mole%). The flow rate of H2 in the second stream is equal to 200 mol/s. This process will include reaction and separation, and two streams will exit the process. The first output stream contains only liquid benzene (product) and toluene (unconverted reactant). The second output stream contains gaseous H2 and CH4. If the conversion of toluene in the process is 75%, determine the following: a) the molar flow rate and mass fraction of benzene in the liquid output stream b) the mole fraction and mass fraction of CH4 in the output gas stream?

Question-AnswerCategory: Chemical ThermodynamicsQ3.7 For the combustion of methane presented in Example 5.4 (of the recommended text – Solen & Harb), the chemical reaction is CH4 +202 → CO2 + 2H2O Let's suppose that methane flows into a burner at 30 mol/s, while oxygen flows into the same burner at 75 mol/s. a) Which reactant is limiting? b) What is the percentage excess of the other reactant? C) If all of the methane is burned, and a single output stream leaves the burner, what is the mole fraction of CO2 in that output stream? [Adapted from Chapter 5, Question 13 of Solen & Harb] Q3.8 A chemical process is used to convert toluene (C7H8) and hydrogen to benzene (C6H6) and methane by the following reaction: C7H3 + H2 → C6H6 + CH4 Two streams enter the process. The first input stream is pure liquid toluene, which enters at a rate of 40 mol/s. The second input stream is a mixture of H2 (95 mole%) and CH4 (5 mole%). The flow rate of H2 in the second stream is equal to 200 mol/s. This process will include reaction and separation, and two streams will exit the process. The first output stream contains only liquid benzene (product) and toluene (unconverted reactant). The second output stream contains gaseous H2 and CH4. If the conversion of toluene in the process is 75%, determine the following: a) the molar flow rate and mass fraction of benzene in the liquid output stream b) the mole fraction and mass fraction of CH4 in the output gas stream? Show transcribed image text Q3.7 For the combustion of methane presented in Example 5.4 (of the recommended text – Solen & Harb), the chemical reaction is CH4 +202 → CO2 + 2H2O Let's suppose that methane flows into a burner at 30 mol/s, while oxygen flows into the same burner at 75 mol/s. a) Which reactant is limiting? b) What is the percentage excess of the other reactant? C) If all of the methane is burned, and a single output stream leaves the burner, what is the mole fraction of CO2 in that output stream? [Adapted from Chapter 5, Question 13 of Solen & Harb] Q3.8 A chemical process is used to convert toluene (C7H8) and hydrogen to benzene (C6H6) and methane by the following reaction: C7H3 + H2 → C6H6 + CH4 Two streams enter the process. The first input stream is pure liquid toluene, which enters at a rate of 40 mol/s. The second input stream is a mixture of H2 (95 mole%) and CH4 (5 mole%). The flow rate of H2 in the second stream is equal to 200 mol/s. This process will include reaction and separation, and two streams will exit the process. The first output stream contains only liquid benzene (product) and toluene (unconverted reactant). The second output stream contains gaseous H2 and CH4. If the conversion of toluene in the process is 75%, determine the following: a) the molar flow rate and mass fraction of benzene in the liquid output stream b) the mole fraction and mass fraction of CH4 in the output gas stream?
Ch asked 1 year ago

Q3.7 For the combustion of methane presented in Example 5.4 (of the recommended text - Solen & Harb), the chemical reaction i
 

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Mazurek Gravity answered 1 year ago

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3.7 CH4 + 207 → cor + 2 Ho Feed: 3 mot City; 75 m or (a) from stichiometry, imole CH4 → 2 mole or 30 mal CH4 + 30xd = 6o molCOV = 0 + 1x30 = 30 mol Sec out in Hyo = nietą € 60 mol o + 2 x 30 - out tncor + + Total moles of product stream = now - 15moles of Cq Hg converted = 40 x 0.75 = 30 mol/sec from stichiomety, Imole Cq Hq can generate Imole C6H6 30 mdo GHz > 30 md 6tTotul mass flow rate of liq stream 1 mol = 30 mogli che x 782, + 10 ml 5 ha * fed home = 2340 Echo + 920 g 5H8 i = 3260 965 2to mw of CH4 = 12+ 4x1= 168/nd mw of Hv = 2x1= 2 g/mol Total mass flow rate of gas stream = 40.5a quel che x 162m + 17ood , X

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