What is the formula for the precipitate formed when solutions of magnesium chloride and potassium phosphate are combined?

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 MgCl2,

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 is 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

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