Sb2S3 + 3Fe --> 2Sb + 3FeS When 0.022mol Sb2S3 react with an excess of Fe, 0.08mol Sb is produced. What is the percent yield of
this reaction? (hint: what's the theoretical yield?)

The actual value is greater than the theoretical value

Given

Reaction

Sb2S3 + 3Fe --> 2Sb + 3FeS

0.022 mol Sb2S3

0.08mol Sb

Required

The percent yield

Solution

Percent yield is the comparison of the amount of product obtained from a reaction(actual) with the amount you calculated (theoretical)

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

Theoretical yield of Sb : 2/1 x 0.022 mol = 0.044 moles

In the question, there seems to be an error, because the actual yield value(0.08=?) should be smaller than the theoretical yield(0.044)