# If the actual yield of PBr_3 s found to be 22.3 g, how do you find the percent yield in this reaction?

## The reaction between phosphorus and liquid bromine is outlined as: $2 P \left(s\right) + 3 B {r}_{2} \left(l\right) \to 2 P B {r}_{3} \left(l\right)$

Dec 3, 2016

You cannot assess the percentage yield in this reaction.

#### Explanation:

$P + \frac{3}{2} B {r}_{2} \rightarrow P B {r}_{3}$

You have reported a mass of $\text{phosphorus bromide}$ that is recovered. There are no data with respect to the amount of bromine or phosphorus used to synthesize the tribromide.

$\text{Moles of}$ PBr_3=(22.3*g)/(270.69*g*mol^-1)=??mol.

So, you know the minimum quantities of bromine and phosphorus used in the synthesis reaction. You do not know the actual quantities utilized.