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2 August, 19:30

How many grams of zinc phosphate are formed when 48.1 mL of 2.18 M zinc nitrate reacts with excess potassium phosphate?

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  1. 2 August, 19:35
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    15.4 g of Zn₃ (PO₄) ₂ are produced

    Explanation:

    Given dа ta:

    Mass of zinc phosphate formed = ?

    Volume of zinc nitrate = 48.1 mL (0.05 L)

    Molarity of zinc nitrate = 2.18 M

    Solution:

    Chemical equation:

    3Zn (NO₃) ₂ + 2K₃PO₄ → Zn₃ (PO₄) ₂ + 6KNO₃

    Moles of zinc nitrate:

    Molarity = number of moles / volume in litter

    Number of moles = 2.18 M * 0.05 L

    Number of moles = 0.109 mol

    Now we will compare the moles of zinc phosphate with zinc nitrate from balanced chemical equation:

    Zn (NO₃) ₂ : Zn₃ (PO₄) ₂

    3 : 1

    0.109 : 1/3*0.109 = 0.04 mol

    0.04 moles of Zn₃ (PO₄) ₂ are produced.

    Mass of Zn₃ (PO₄) ₂:

    Mass = number of moles * molar mass

    Mass = 0.04 mol * 386.1 g/mol

    Mass = 15.4 g
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