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13 August, 10:18

121 grams of iron (II) acetate reacts with 360 grams of lead (II) oxalate. Predict the products, write the complete reaction equation, balance, and calculate the mass of the precipitate.

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  1. 13 August, 13:08
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    The mass of the precipitate is 100.2 grams

    Explanation:

    Step 1: Data given

    Mass of iron (II) acetate = 121.0 grams

    Mass of lead (II) oxalate = 360.0 grams

    Molar mass Fe (C2H3O2) 2 = 173.93 g/mol

    Molar mass PbC2O4 = 295.22 g/mol

    Step 2: The balanced equation

    Fe (C2H3O2) 2 + PbC2O4 → FeC2O4 + Pb (C2H3O2) 2

    Step 3: Calculate moles iron (II) acetate

    Moles = mass / molar mass

    Moles = 121.0 grams / 173.93 g/mol

    Moles = 0.696 moles

    Step 4: Calculate moles lead (II) oxalate

    Moles = 360.0 grams / 295.22 g/mol

    Moles = 1.219 moles

    Step 5: Calculate limiting reactant

    For 1 mol iron (II) acetate we need 1 mol lead (II) oxalate to produce 1 mol Iron (II) oxalate and 1 mol Lead (II) acetate

    iron (II) acetate is the limiting reactant. It will completely be consumed (0.696 moles). Lead (II) oxalate is in excess. There will react 0.696 moles. There will remain 1.219 - 0.696 = 0.523 moles

    Step 6: Calculate moles of iron (II) oxalate

    For 1 mol iron (II) acetate we need 1 mol lead (II) oxalate to produce 1 mol Iron (II) oxalate and 1 mol Lead (II) acetate

    For 0.696 moles iron (II) acetate we'll have 0.696 moles iron (II) oxalate

    Step 7: Calculate mass iron (II) oxalate

    Mass = moles * molar mass

    Mass = 0.696 moles * 143.91 g/mol

    Mass = 100.2 grams

    The mass of the precipitate is 100.2 grams
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