Asked by Gianna

When 20.1 g Sb2S3 reacts with an excess of Fe, 9.84 g Sb is produced. What is the percent yield of this reaction?

My work so far...

converted 20.1g of Sb2S3 to moles
20.1/149.8=.134

.134Sb2S3*2moles Sb/1 mole Sb2S3=.268*121.7=32.6 of Sb <---Thoretical yeild.

9.84/38.6= .3018*100= 30.18%

I have a webassign which tells you if the answer is correct or not and apparently it said mine wasn't
Answered by DrBob222
You didn't write the equation but it appears you used Fe2S3 as the product. I think perhaps it is FeS that forms
Sb2S3 + 3Fe ==> 2Sb + 3FeS
Try that and see if it is any better. Your process looks ok if you started with the right equation.
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