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26 February, 13:58

2AgNO3 (s) + MgCl2 (aq) - -> 2AgCl (s) + Mg (NO3) 2 (aq) The student starts with 50.0 grams of mgcl2 and an excess of AgNO3. If she makes 125.1g of AgCl, what is her percent yield?

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  1. 26 February, 16:37
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    The percent yield is 83.12%

    calculation

    percent yield = actual yield / theoretical yield x 100

    actual yield = 125.1 g

    Theoretical yield is calculated as below

    Step 1: find the moles of MgCl₂

    moles = mass: molar mass

    from periodic table the molar mass of mgCl₂ = 24.3 + (35.5 x2) = 95.3 g/mol

    moles = 50.0 g: 95.3 g/mol = 0.525 moles

    Step 2: use the mole ratio to determine the moles of AgCl

    from given equation MgCl₂: AgCl is 1:2

    therefore the moles of AgCl = 0.525 x2/1 = 1.05 moles

    Step 3: find the theoretical mass AgCl

    mass = moles x molar mass

    from periodic table the molar mass of Agcl = 107.87 + 35.5 = 143.37 g/mol

    mass = 1.05 moles x 143.37 g/mol = 150.5 grams

    The percent yield is therefore = (125.1 g/150.5) x 100 = 83.12%
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