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 mol% C₂H₅Br and 12.3 mol% HBr.

a) Calculate the moles of ethyl bromide in the product stream.

b) Determine the mole fraction of hydrogen bromide in the product.

c) Find the percentage yield of the reaction.

Without the total number of moles provided, we can only describe methods to calculate the moles of ethyl bromide and the mole fraction of HBr in the product stream, and to find the percentage yield, theoretical and actual yields are required.

Explanation:

Calculating Moles of Ethyl Bromide and Mole Fraction

To calculate the moles of ethyl bromide (C2H5Br) in the product stream, you would need to know the total number of moles in the stream. Unfortunately, that information is not provided here. Typically, you would multiply the total number of moles by the mole percentage of ethyl bromide to find the exact amount. However, we can discuss the method assuming an arbitrary total number of moles. For instance, if there were 100 total moles in the stream, then there would be 51.7 moles of ethyl bromide because it makes up 51.7% of the total.

To determine the mole fraction of hydrogen bromide (HBr), you divide the moles of HBr by the total number of moles in the product stream. Again, without knowing the total number of moles, we use the percentage directly. The mole fraction is 12.3% or 0.123 if we assume the total mole percent adds up to 100%.

The percentage yield of the reaction cannot be calculated accurately without knowing the initial amounts of reactants or the theoretical yield. If given the theoretical yield, the percentage yield is calculated by taking the actual yield (moles of ethyl bromide produced) and dividing by the theoretical yield, then multiplying by 100.