Question

How many grams of calcium phosphate are precipitated when 25.0 mL of .220 M calcium chloride solution are allowed to react with 15.0 mL of .880 M phosphoric acid solution?

Answer 1

g of Ca3(PO4)2

V = 25 ml

M = 0.22 CaCl2

V = 15 ml

M = 0.88 H3PO4

the reaction

CaCl2 + H3PO4 = ?

The reaction is:

H3PO4+CaCl2=Ca3(PO4)2+HCl

when balanced:

2 H3PO4 + 3 CaCl2 = Ca3(PO4)2 + 6 HCl

find limiting reactants

mmol of H3PO4 = M2*V2 = 15*0.88 = 13.2

mmol of CaCl2 = M1*V1 = 25*0.22 = 5.5

the ratio is 2 mol of H3PO4 per 3 mol of CaCl2 or 1:1.5

there is plenty of acid, therefore CaCl2 is limiting

5.5 mmol of CaCl2 will form 1/3 of product

that is

5.5/3 = 1.8333 mmol of Ca3PO42 is expected to precipitate

mass = mol*MW = (1.8333 *10^-3 ) * 310.1767 = 0.5685966 g of preciitate