The reaction between ethylene and hydrogen bromide to form ethyl bromide is carr

Question

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 59.7 mol% C2H5Br and 19.3 mol% HBr.

The feed to the reactor conains only ethylene and hydrogen bromide.

Calculate the fractional conversion of the 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?

i just need the last question

Explanation / Answer

Here in this HBr is limiting reagent

% excess C2H4 = (actual - stoichiometric) / actual x 100

Actual feed for C2H4 = 165 x 0.545 = 89.93 mol/s

Theoritical requirement for C2H4 based on stoicchiometry:

165 [(1-0.545) HBr ] [1 mol C2H4 / 1 mol HBr ] = 75.08 mol/s

% excess C2H4 = (89.93 - 75.08) / 75.08 x 100 = 19.8%

Fractional conversion of HBr = amount reacted / amount fed = input - output / input

= (165) ( 1 - 0.545) - ( 108.77 - 0.173) / (165) (1 -  0.545)

= 0.749

The extent of reaction E can be determined based on C2H4 , HBr, C2H5Br:

C2H4 : 0.310(108.77) = (165)(0.545) – E

HBr: 0.173(108.77) = (165)(1-0.545) – E

C2H5Br: 0.517(108.77) = 0 – E

Solving for E: E = 56.2 mol/s

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