One way to synthesize ethylamine (CH3CH2NH2) is from the reaction of ammonia (NH

Question

One way to synthesize ethylamine (CH3CH2NH2) is from the reaction of ammonia (NH3) with chloroethane (CH3CH2CI) NH-CH,CH.cl CH3CH.NH-HCI One problem with this synthesis route is that the above reaction is not very selective, and ammonia may react with two chloroethane molecules to form diethylamine ((CH3CH2)2NH) NH,+2CH,CH,CI (CH,CH,),NH+2HCl A mixture of 0.465 mol NH3/mol, 0.465 mol CH3CH2Cl/mol and the remainder inerts is fed into a reactor Within the reactor, the fractional conversion of CH3CH2Cl is 0.840 and the fractional yield of CH3CH2NH2 is 0.530. [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 Reactor X1 = 0.465 mol NH/mol x2 0.465 mol CH3CH2Cl/mol x3 mol inerts/mol n4 mol NH3 n5 mol CH3CH2CI ng mol inerts n mol CH3CH2NH2Scroll down to ns mol (CH3CH2)2NH for the nine n mol HI enter in values

Explanation / Answer

Moles of NH3 enter the reactor = moles of feed x mol fraction of NH3

n1 = 100 mol x 0.465 = 46.5 mol

Moles of CH3CH2Cl enter = moles of feed x mol fraction of CH3CH2Cl

n2 = 100 mol x 0.465 = 46.5 mol

Moles of inerts enter = moles of feed x mol fraction of inerts

n3 = 100 mol x (1-0.465-0.465) = 7 mol

Moles of inerts leave = moles of inerts enters

n6 = 7 mol

Since the inerts do not participate in the reaction

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