Use the following equation to answer the question below:
sb2s3(s) + 3fe(s) --> 2sb(s) + 3fes(s)
when 18.45g sb2s3 reacts with an excess of fe, 12.1g sb is produced. what is the percent yield of this reaction?
65.6%
75.4%
84.1%
91.5%

1) Balanced chemical equation

Sb2S3(s) + 3Fe(s) --> 2Sb(s) + 3FeS(s)

2) Theoretical molar ratios

1 mol SbsS3 : 3 mol Fe : 2 mol Sb : 3 mol FeS

3) Convert 18.45g Sb2S3 and12.1g Sb to moles

You need to use the atomic and molar masses

Sb: 121.76 g/mol

S: 32.07 g/mol

Sb2S3: 2 * 121.76 g/mol + 3 * 32.07 g/mol = 339.73 g/mol

18.45 g Sb2S3 / 339.73 g/mol = 0.0543 mol Sb2S3

12.1 g Sb / 121.76 g / mol = 0.099 mol Sb

Theoretical yield:

1 mol Sb2S3 / 2 mol Sb = 0.0543 mol SbsS3 / x mol Sb

Solve for x:

x = 0.0543 mol Sb2S3 * 2 mol Sb / 1 mol Sb2S3 = 0.1086 mol Sb

4) Percent yield

0.099 mol Sb / 0.1086 mol Sb * 100 = 91.16%

91.5% (the difference with 91.16% is due to the decimals used for the atomic masses)

Sorry i dont understand the question can you rephase it?