<span>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 </span><span>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%
Answer: 91.5% (the difference with 91.16% is due to the decimals used for the atomic masses)
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Anymore it would be technical logically
