Question

A 75.0 mL solution of 0.240 M Na3PO4 is combined with excess CaCl2.

1. How many grams of calcium phosphate will form as a precipitate if all of the Na3PO4 is consumed in the reaction?

2. What is the percent yield for the reaction if 2.13 g of calcium phosphate is produced?

Answer 1

1)

moles of Na33PO4 reacting = M(Na3PO4) * V(Na3PO4)

= 0.240 M * 0.0750 L

= 0.0180 mol

The reaction is:

2 Na3PO4 + 3 CaCl2 —> Ca3(PO4)2 + 6 NaCl

From reaction,

mol of Ca3(PO4)2 formed = (1/2)*number of moles of Na3PO4 reacting

= (1/2)*0.0180 mol

= 9.0*10^-3 mol

Molar mass of Ca3(PO4)2,

MM = 3*MM(Ca) + 2*MM(P) + 8*MM(O)

= 3*40.08 + 2*30.97 + 8*16.0

= 310.18 g/mol

use:

mass of Ca3(PO4)2,

m = number of mol * molar mass

= 9*10^-3 mol * 3.102*10^2 g/mol

= 2.792 g

Answer: 2.79 g

2)

% yield = actual yield * 100 / theoretical yield

= 2.13*100/2.79

= 76.3 %

Answer: 76.3 %