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26 September, 02:33

A 500-g sample of Al2 (SO4) 3 is reacted with 450 g of Ca (OH) 2. A total of 596 g of calcium sulfate is produced in the laboratory. What is the percent yield?

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  1. 26 September, 02:52
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    The percent yield is 99.9 %

    Explanation:

    Step 1: Data given

    Mass of a sample Al2 (SO4) 3 = 500 grams

    Mass of Ca (OH) 2 = 450 grams

    Mass of CaSO4 produced = 596 grams

    Molar mass Al2 (SO4) 3 = 342.15 g/mol

    Molar mass Ca (OH) 2 = 74.09 g/mol

    Step 2: The balanced equation

    Al2 (SO4) 3 + 3Ca (OH) 2 → 2Al (OH) 3 + 3CaSO4

    Step 3: Calculate moles

    Moles = mass / molar mass

    Moles Al2 (SO4) 3 = 500 grams / 342.15 g/mol

    Moles Al2 (SO4) 3 = 1.46 moles

    Moles Ca (OH) 2 = 450 grams / 74.09 g/mol

    Moles Ca (OH) 2 = 6.07 moles

    Step 4: Calculate the limiting reactant

    For 1 mol Al2 (SO4) 3 we need 3 moles Ca (OH) 2 to produce 2 moles Al (OH) 3 and 3 moles CaSO4

    Al2 (SO4) 3 is the limiting reactant. It will completely be consumed (1.46 moles). Ca (OH) 2 is in excess. There will react 3*1.46 = 4.38 moles

    There will remain 6.07 - 4.38 = 1.69 moles

    Step 5: Calculate moles CaSO4

    For 1 mol Al2 (SO4) 3 we need 3 moles Ca (OH) 2 to produce 2 moles Al (OH) 3 and 3 moles CaSO4

    For 1.46 moles Al2 (SO4) 3 we'll have 3*1.46 moles = 4.38 moles CaSO4

    Step 6: Calculate mass CaSO4

    Mass CaSO4 = 4.38 moles * 136.14 g/mol

    Mass CaSO4 = 596.3 grams

    Step 7: Calculate percent yield

    Percent yield = (actual yield / theoretical yield) * 100%

    Percent yield = (596 grams / 596.3 grams) * 100 %

    Percent yield = 99.9 %

    The percent yield is 99.9 %
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