Answer:
See explanation below
Explanation:
To calculate any percent yield of a reaction you need to provide, the theorical data and the experimental data, which you are not providing.
I will do an example with some values I found on another place to explain to you how to do it. You then, replace the data you have and follow the same procedure.
Let's suppose we have the following data of the stilbene:
density: 1.0111 g/mL
volume used: 0.3 mL
Molecular weight: 180.25 g/mol
Now, we use 0.3 mL of cis stilbene to do a reaction with acid and bromine to produce the 1,2-dibromo-1,2-diphenylethane.
The problem stated that the cis stilbene is the limiting reactant, therefore, the moles consumed of stilbene, would be the moles produced of the final product.
With the density let's calculate the mass of stilbene, and then, with the molecular weight, the moles:
d = m/V ---> m = d*V
m = 1.011 * 0.3 = 0.3033 g
moles = 0.3033 / 180.25 = 0.0017 moles
These obtained moles would be the moles of the final product too, because stilbene is the limiting reactant so:
moles of product = 0.0017 moles
Let's calculate the mass:
Molecular weight of 1,2-dibromo-1,2-diphenylethane = 339.8 g/mol
m = 0.0017 * 339.8 = 0.5776 g
This would be the theorical mass obtained in the experiment. Now, let's suppose we obtained a mass of 0.4158 g. This is the actual yield of the reaction, so the percetn yield would be:
%yield = Exp yield / theo yield * 100
Replacing:
%yield = 0.4158/0.5776 * 100
<em>%yield = 71.99 %</em>
This would be the %yield of the bromination. All you have to do now, is replace your theorical and experimental data and you should get to the final and accurate yield.
