Question

14. When a solution of magnesium chloride (MgCl2) is poured into a solution of potassium phosphate (K3PO4) a precipitate forms.

a. Write the chemical equation for this reaction and identify the precipitate.

b. Write the net ionic equation for this reaction.

c. If 30.0 mL of 0.250 M magnesium chloride and 13.0 mL of 0.500 M potassium phosphate are mixed, how many grams of precipitate can be formed?

Answer 1

a. Chemical equation for reaction is:

3 MgCl₂(aq) + 2 K₃PO₄(aq) → Mg₃(PO₄)₂(s) + 6 KCl(aq)

The precipitate formed is Magnesium phosphate, Mg₃(PO₄)₂.

b. Net ionic equation:

Ionic equation is 3Mg²⁺ (aq) + 6Cl^- (aq) + 6K⁺ (aq) + 2(PO₄)^3- (aq) → 6K⁺ (aq) + 6Cl^-(aq) + Mg₃(PO₄)₂ (s)

After cancelling out spectator ions, we get the net ionic equation as:

3Mg²⁺ (aq) + + 2(PO₄)^3- (aq) → Mg₃(PO₄)₂ (s)

c. Reaction is 3 MgCl₂(aq) + 2 K₃PO₄(aq) → Mg₃(PO₄)₂(s) + 6 KCl(aq)

MgCl2 → 3 mol, 30 ml, 0.25 M

  K₃PO₄ →​ 2 mol, 13 ml, 0.5 M

Calculate moles of reactants,

Moles of MgCl₂ = 30ml * 0.25 mol / 1000 ml = 7.5 x 10^-3 mol

Moles of K₃PO₄ = 13 ml * 0.5 mol / 1000 ml = 6.5 x 10^-3 mol

Determining the limiting reactant,

For MgCl_2,

3 mol of MgCl₂ gives 1 mol of Mg₃(PO₄)₂

Therefore, 7.5 x 10^-3 mol of MgCl₂ gives 2.5 x 10^-3 mol of Mg₃(PO₄)_2.

For K₃PO₄ ,

2 mol of K₃PO₄ gives 1 mol of Mg₃(PO₄)₂

Therefore, 6.5 x 10^-3 mol of K₃PO₄ gives 3.25 x 10^-3 mol of Mg₃(PO₄)_2.

Therefore, MgCl2 is the limiting reactant.

Hence, 2.5 x 10^-3 mol of Mg₃(PO₄)_{2 i}s produced.

To convert to grams,

Mass = Number of moles * Molar mass

= 2.5 x 10^-3 * 262.86

= 0.657 g of Mg₃(PO₄)₂