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The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. The product stream is analyzed and found to contain 51.7 mole% C H-Br and 17.3% HBr. The feed to the reactor contains only ethylene an h drogen bromide Calc te era tu a converso e limiting reactant and the percentage by which the other reactant is in excess. If the molar flow rate of the feed stream is 165 mol/s, what is the extent of reaction?

Explanation / Answer

C2H4 + HBr     --------------------> C2H5Br

mol % of C2H5Br = 51.7 ,

mol % of HBr ( unreacted) = 17.3

mol % of C2H4 (unreacted) = 100 - 51.7 -17.3 = 31

total HBr mol % initially = 51.7+17.3 = 69

total C2H4 initially = 31 + 51.7 = 82.7

since C2H4 and HBr react in 1:1 ration

limiting reagent is HBr , C2H4 is excess

fraction of conversion of limiting reganet =( 51.7/69) = 0.7493

% of excess reagent in excess = ( 82.7-69) x 100 / 69 = 19.86 %

% of excess reagent in excess = 20 % nearly.

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