Problem 5.50 The reaction C2H4+HBrC2H5Br is carried out in a continuous reactor.
ID: 984912 • Letter: P
Question
Problem 5.50 The reaction C2H4+HBrC2H5Br is carried out in a continuous reactor. The product stream is analyzed, and is found to contain 50 mole % C2H,Br and 33.3% HBr. The feed to the reactor contains only C2H4 and HBr. Calculate the fractional conversion of the limiting reactant and the percentage by which the other reactant is in excess. In order to increase the conversion a splitter is introduced after the reactor, which feeds 20% of the reaction product back to mix with the incoming stream of reactants. Calculate the overall conversion.Explanation / Answer
Xaout=0.5%, C2H5Br
Xbout=0.33% HBR
The rest of the product would be the C2H4 without reaction
Xcout=0.17%
given taht C2H4 and C2H5Br have a 1:1 reaction relation we can say that every C2H5Br mole produced is because a mole of C2H4 reacted
Taking a base as reference of 100mol in the output the distribution between the compouds would be
C2H5Br=50moles
HBr=33moles
C2H4=17 moles
in the entry we can say that there is a total of 67 moles of C2H4 and 88moles of HBR
fractional coversion of C2H4 (limitant agent) =(67-17)/67=0.75
HBr excess percentage is (88-50)/50*100=76%