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)

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i