One way to synthesize ethylamine (CH₂CH₂NH₂) is from the reaction of ammonia (NH3) with chloroethane (CH₂CH₂CI). (1) NH₂ + CH₂CH₂Cl → CH₂CH₂NH₂ + HCl One problem with this synthesis route is that the reaction is not very selective, and ammonia may react with two chloroethane molecules to form diethylamine ((CH3 CH₂)₂NH). (2) NH3 + 2CH₂CH₂Cl → (CH₂CH₂)₂NH + 2HCl A mixture of x₁ = 0.465 mol NH3/mol, x₂ = 0.465 mol CH3CH₂Cl/mol, and the remainder inerts is fed into a reactor. Within the reactor, the fractional conversion of CH₂ CH₂Cl is 0.750 and the fractional yield of CH₂CH₂NH₂ is 0.410. [Fractional yield is the (moles of desired product formed)/(moles of product possible for complete conversion of the feed to the desired product).] Assume a 100 mol basis for the feed stream, and calculate the number of moles of each component entering the reactor and leaving in the product stream. Here is the process flow diagram for reference. 100 mol x, mol NH₂ /mol x₂ mol CH₂CH₂Cl/mol x₂ mol inerts/mol Reactor mol NH3 mol CH₂CH₂CI Attempt 2 n4 ns nmol inerts n, mol CH,CH,NH, ng mol (CH,CH,),NH n, mol HCI

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
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Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Question
How many moles of NH3 leave the reactor?
How many moles of HCl leave the reactor?
n4=
ng =
13.95
37.9
mol NH3
mol HCI
Transcribed Image Text:How many moles of NH3 leave the reactor? How many moles of HCl leave the reactor? n4= ng = 13.95 37.9 mol NH3 mol HCI
Macmillan Learning
One way to synthesize ethylamine (CH₂CH₂NH₂) is from the reaction of ammonia (NH3) with chloroethane (CH₂CH₂CI).
(1) NH3 + CH₂CH₂Cl → CH₂CH₂NH₂ + HCl
One problem with this synthesis route is that the reaction is not very selective, and ammonia may react with two chloroethane
molecules to form diethylamine ((CH₂CH₂)₂NH).
(2) NH3 + 2 CH₂CH₂Cl
(CH₂CH₂)₂NH + 2HCl
A mixture of x₁ = 0.465 mol NH3/mol, x2 = 0.465 mol CH₂ CH₂Cl/mol, and the remainder inerts is fed into a reactor. Within
the reactor, the fractional conversion of CH3 CH₂Cl is 0.750 and the fractional yield of CH₂ CH₂NH₂ is 0.410. [Fractional yield
is the (moles of desired product formed)/(moles of product possible for complete conversion of the feed to the desired product).]
Assume a 100 mol basis for the feed stream, and calculate the number of moles of each component entering the reactor and
leaving in the product stream. Here is the process flow diagram for reference.
100 mol
mol NH3 /mol
X₁
x₂ mol CH₂CH₂Cl/mol
x₂ mol inerts/mol
Reactor
nạ mol NH,
n mol CH₂CH₂CI
mol inerts
n6
n₂
mol | CH,CH,NH,
, mol (CH,CH,),NH
n8
n, mol HCI
Attempt 20
Transcribed Image Text:Macmillan Learning One way to synthesize ethylamine (CH₂CH₂NH₂) is from the reaction of ammonia (NH3) with chloroethane (CH₂CH₂CI). (1) NH3 + CH₂CH₂Cl → CH₂CH₂NH₂ + HCl One problem with this synthesis route is that the reaction is not very selective, and ammonia may react with two chloroethane molecules to form diethylamine ((CH₂CH₂)₂NH). (2) NH3 + 2 CH₂CH₂Cl (CH₂CH₂)₂NH + 2HCl A mixture of x₁ = 0.465 mol NH3/mol, x2 = 0.465 mol CH₂ CH₂Cl/mol, and the remainder inerts is fed into a reactor. Within the reactor, the fractional conversion of CH3 CH₂Cl is 0.750 and the fractional yield of CH₂ CH₂NH₂ is 0.410. [Fractional yield is the (moles of desired product formed)/(moles of product possible for complete conversion of the feed to the desired product).] Assume a 100 mol basis for the feed stream, and calculate the number of moles of each component entering the reactor and leaving in the product stream. Here is the process flow diagram for reference. 100 mol mol NH3 /mol X₁ x₂ mol CH₂CH₂Cl/mol x₂ mol inerts/mol Reactor nạ mol NH, n mol CH₂CH₂CI mol inerts n6 n₂ mol | CH,CH,NH, , mol (CH,CH,),NH n8 n, mol HCI Attempt 20
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