Finding the percent yield? Help!

Pure antimony (Sb) can be prepared by heating the antimony ore (Sb2S3) in the presence of iron as follows:
Sb2S3 + 3Fe ==> 2Sb + 3FeS

If 353.6 g of antimony sulfide is reacted with iron and 24.9 g of pure antimony is produced, what is the % yield for this reaction?

Use stoichiometery to first determine the theoretical amount of Sb that could be produced. Then divide the yield of antimony by the theoretical yield and multiply by 100 to convert the fraction to a percent.

(356-g Sb2S3 / 339.9 g/mol Sb2S3( x (2 moles Sb / 1 mole Sb2S3) x (121.8 g Sb / 1 mole Sb) = 255 g Sb

24.9 g Sb / 255 g Sb x 100% = 9.76%

Are you absolutely sure that only 24.9g of pure antimony is produced? Seems like that would be a very small percent yield (around 9%).