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

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.