Antimony (iii) sulfide is reacted with excess iron. if 175.6 grams of pure antimony is produced, what mass of antimony (iii) sulfide was required to start with? sb2s3(s) + 3fe(s)→2sb(s) +3fes(s)

The balanced chemical reaction is:

Sb2S3(s) + 3Fe(s) → 2Sb(s) +3FeS(s)

We are given the amount of pure antimony that is to be produced from the reaction. Since the amount of iron is in excess, then what is important is the amount of the antimony sulfide that is to be supplied to produce the given amount of antimony. We relate the substances by the use of the reaction after converting the mass to units of moles. We calculate as follows:

175.6 g Sb ( 1 mol Sb / 121.76 g Sb ) ( 1 mol Sb2S3 / 2 mol Sb ) ( 339. 37 g / mol ) = 244.95 g Sb2S3

Therefore, approximately 245 g of Sb2S3 should be supplied to the reaction.

Hi + naoh -> h2o + nal

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