2-Fluoropentane can be dehalogenated through a series of two reactions. The firs

Question

2-Fluoropentane can be dehalogenated through a series of two reactions. The first reaction occurs with 20.0% excess sodium hydroxide within the solvent methanol, and produces two alkenes; 70% is 1-pentene and 30% is 2-pentene. The first reaction is shown below. 100% of the 2-fluoropentane reacts to form one or the other pentene product. The product stream proceeds to a separator that removes the methanol, sodium fluoride, water and unreacted sodium hydroxide. In the second reactor, 20.0% excess hydrogen gas is added to the pentene stream, and 70.0% of each pentene is converted to pentane. Given the feed concentrations in the flow diagram below, if 20.0 mol/hr of pentane is required, what are the flow rates of each feed and product stream? q1, q2, q3, q4, q5?

Explanation / Answer

pentane = 20mol/hr
This is only 70% of pentene
Total pentene flow = 100*20/70 = 28.6 mol/hr
rate of h2 = 120% of 28.6 = 34.3 mol /hr
so,
q4 = 34.3 mol /hr

q5 = 20 mol/hr pentane + unreacted pentene + excess H2
     =20 + (28.6-20) + (34.3-20)
     = 42.9 mol/hr

amount of pentene = 28.6mol/hr = rate of NaF = rate of H2O
rate of NaOH = 120% of 28.6 = 34.3 mol/hr
rate of CH3OH = 9*rate of C5H11F + 19*rate of NaOH
                               =9*rate of pentene + 19*rate of NaOH
                               = 9*28.6 + 19*34.3
                               =909.1 mol/hr
q3 = 909.1 + 34.3 + 28.6+28.6 = 1000.6mol/hr

q1 = 28.6 + 9*28.6 = 286mol/hr

q2 =34.3 * 19*34.3 = 686mol /hr

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