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 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