Can you please show the work?Thanks. 20,0% excess sodium hydroxide within the so

Question

Can you please show the work?Thanks.

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. CH, CFHCH2CH,CH, CH, OH+ NaOH CH CHCH,CH,CH, 70% CH,C CHCH,CH, 0% CH, OH NaF+ Hro 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 a second reactor, 20.0% excess hydrogen gas is added to the pentene stream, and 70.0% of each pentene is converted to pentane CH CHCH, CH,C H C H,2 CH,CH- CHCH CH H. CSH12 Given the feed concentrations in the flow diagram below, if 20.0 molhr of pentane is required, what are the flow rates of each feed and product stream? mol/hr mol H2hr 0.95 mol CH3OH/mol 0.05 mol NaOH/mol Reactor Separator Reactor qs mol/hr 0.10 mol CSHffF/mol 0.90 mol CH CH3OH 1-pentene NaF 2-pentene H20 NaOH mol/hr mol /hr mol /hr mol/hr mol hr

Explanation / Answer

Assuming that 50% of petane formed is obtained by each 1-pentene and 2-pentene then from the two reactions the feed input should be inthe ratio of 1:2.since all the components are reacting in 1: 1 molar proportions then feed input from reaction one(70%1-penteneand 30%2-pentene) and two (100% of either of 1-pentene or 2-pentene) should be in the ratio of 1:2.

from the reaction one : since 0.05mol of 2flouropentane is present in the reaction mixture (calculated from amount of excess amount of sodium hydroxide)

70% of it is converted to 1-pentene = 0.035moles and the rest 0.015 moles is 2-pentene

from the second reaction any one type of pentene is 100% produced

so the total 0.1 moles is converted to either 1-pentene or 2-pentene.(assuming each to be produced one at a time)

if only 1-pentene is produced from reaction two then : the feed to the second reactor consists of 0.135moles 1-pentene and 0.015moles 2-pentene

70% of each is converted to pentane in the final reaction

70% of 0.135 moles of 1-pentene = 0.0945moles of pentane

70% of 0.015moles of 2-pentene = 0.0105 moles of pentane

hence total of 0.1mole of pentane is produced.

and if only 2-pentene is formed then the feed to the second reactor consists of 0.035moles of 1-pentene and 0.115moles of 2-pentene

70% of 0.035 moles of 1-pentene = 0.0245moles of pentane

70% 0f 0.115 moles of 2-pentene = 0.0805moles of pentane.

put together equal to 0.1 moles again.

so either way 1/3rd of first reaction and 2/3 of second reaction is required for producing each of 10 moles of pentane

then 1/3rd of 10 is3.334moles but required pentane is 20moles so it is 6.008moles

and 2/3 of 10 is 6.666moles but for 20 moles it is13.332moles

therefore q2=3.334moles/hr

q1=13.332moles/hr

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