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11 March, 18:20

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%

+5
Answers (1)
  1. 11 March, 20:37
    0
    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%

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