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.
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
1 answer
1 answer