Midterm_3_2006 solution - Chemical Engineering 140 Midterm...

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Chemical Engineering 140 November 20th, 2006 Midterm #3 (200) 1. Production of Maleic Anhydride Figure P18.28 (per the text but with changes !) below gives a process flow diagram for the production of maleic anhydride by the partial oxidation of benzene. Pertinent reactions are listed below. The moles of O 2 fed to the reactor per mole of pure benzene fed to the reactor is 10. Vapor leaves the steam still (shown at the bottom left of the diagram) at 180 o F and saturated with both benzene and water before entering the reactor. All of the maleic acid produced in the reactor is removed with water in the bottom stream from the water scrubber. All of the C 6 H 6 , O 2 , CO 2 , and N 2 leaving the reactor leave in the stream from the top of the water scrubber, saturated with H 2 O. Originally, the benzene contains trace amounts of a nonvolatile contaminant that inhibit the reaction. This contaminant is removed by steam distillation in the steam still. The steam still contains liquid phases of both benzene and water (benzene is completely insoluble in water). The benzene phase is 80% by weight, and the water phase is 20% by weight of the total of the two liquid phases in the still. Other process conditions are given in the flow sheet. Use the vapor-pressure data in Table 1. 1 Moist Air 1 atm Temperature = 150 o F 20% R.H,
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Chemical Engineering 140 November 20th, 2006 Midterm #3 Choose 1 mole of pure benzene fed into the reactor as a basis. (10) (a) Calculate the moles of benzene reacted in Reaction (1) using the given basis. Basis = 1 mole of pure Benzene into the reactor Given in the problem, the ratio between mole of O2 fed and mole of Benzene fed is 10. Therefore, mole of Oxygen into the reactor = 10 mole. N2 into the reactor = (10mol O2)*(79 mole N2/21 mole O2) = 37.6 mole of N2 into the reactor. Total mole coming out of the reactor = (37.6 mole of N2)*(1/0.8) = 47 mole on Dry Basis 2
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Chemical Engineering 140 November 20th, 2006 Midterm #3 Mole of C4H4O4 out of the reactor = (0.0076)*(47) = 0.357 mole of C4H4O4 Amount of Benzene reacted in Reaction (1) = (0.357)*(1mole C4H4O4)/(1mole Benzene) =
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This note was uploaded on 09/15/2009 for the course CHEM 1 taught by Professor Pines during the Spring '08 term at University of California, Berkeley.

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Midterm_3_2006 solution - Chemical Engineering 140 Midterm...

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