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18 January, 08:05

A 5.95-g sample of AgNO3 is reacted with BaCl2 according to the equation to give 4.07 g of AgCl. What is the percent yield of AgCl?

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  1. 18 January, 08:16
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    The % yield of AgCl is 81.1 %

    Explanation:

    Step 1: Data given

    Mass of AgNO3 = 5.95 grams

    Molar mass AgNO3 = 169.87 g/mol

    Molar mass BaCl2 = 208.23 g/mol

    Mass of AgCl = 4.07 grams

    Molar mass of AgCl = 143.32 g/mol

    Step 2: The balanced equation

    2AgNO3 + BaCl2 → 2AgCl + Ba (NO3) 2

    Step 3: Calculate moles AgNO3

    Moles AgNO3 = mass AgNO3 / molar mass AgNO3

    Moles AgNO3 = 5.95 grams / 169.87 g/mol

    Moles AgNO3 = 0.0350 moles

    Step 4: Calculate moles AgCl

    For 2 moles AgNO3 we need 1 mol BaCl2 to produce 2 moles AgCl and 1 mol Ba (NO3) 2

    For 0.0350 moles AgNO3 we'll have 0.0350 moles AgCl

    Step 5: Calculate mass AgCl

    Mass AgCl = moles AgCl * moles AgCl

    Mass AgCl = 0.0350 moles * 143.32 g/mol

    Mass AgCl = 5.016 grams

    Step 6: Calculate % yield AgCl

    % AgCl = (4.07 grams / 5.016 grams) * 100 %

    % AgCl = 81.1 %

    The % yield of AgCl is 81.1 %
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