The gas-phase reaction between methanol and acetic acid to form methyl acetate and water takes place in a batch reactor. CH;OH + CH;COOH - CH¿COOCH3 + H2O When the reaction mixture comes to...

Extracted text: The gas-phase reaction between methanol and acetic acid to form methyl acetate and water takes place in a batch reactor. CH;OH + CH;COOH - CH¿COOCH3 + H2O When the reaction mixture comes to equilibrium, the mole fractions of the four reactive species are related by the reaction equilibrium constant: K = YCH,COOCH, "VH,0 YCH,OH "YCH,COOH = 4.87 Part a) If the feed to the reactor contains equimolar quantities of methanol and acetic acid and no other species, calculate the equilibrium fractional conversion as a percentage. Part b) It is desired to produce 70 moles of methyl acetate starting with 75 moles of methanol. How many moles of acetic acid must be fed for there to be 70 moles of methyl acetate at equilibrium? Part c) What is the composition of the final product? Enter as percentages in the order: Methyl Acetate, Water, Methanol, Acetic Acid. Part d) Repeat Part b but determine the moles of acetic acid AND methanol required to achieve 99% conversion of the methanol while also producing 70 moles of methyl acetate. Enter your answer in the order: moles of acetic acid, moles of methanol