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: #2P(s) + 3Br_2(l) -> 2PBr_3(l)#

Answer 1

You cannot assess the percentage yield in this reaction.

#P+3/2Br_2 rarr PBr_3#

You have reported a mass of #"phosphorus bromide"# that is recovered. There are no data with respect to the amount of bromine or phosphorus used to synthesize the tribromide.

#"Moles of"# #PBr_3=(22.3*g)/(270.69*g*mol^-1)=??mol#.

Therefore, while you are aware of the minimum amounts of phosphorus and bromine that were used in the synthesis reaction, you are unaware of the actual amounts that were used.

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Answer 2

Caleb Adams

Percent yield formula:

[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.