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

15.4 g of Zn₃(PO₄)₂ are produced

Explanation:

Given data:

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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