Question

A sample of 3.53 g of K3PO4 is added to 75.3 mL of 0. 545 M AgNO3, resulting in the formation of 4.86 g of a Ag3PO4 precipitate. Calculate the percentage yield of the reaction.

3Ag+(aq)   +    PO4 (aq)   →    Ag3PO4(s)

Answer 1

calculate mol of K3PO4:

Molar mass of K3PO4,

MM = 3*MM(K) + 1*MM(P) + 4*MM(O)

= 3*39.1 + 1*30.97 + 4*16.0

= 212.27 g/mol

mass(K3PO4)= 3.53 g

number of mol of K3PO4,

n = mass of K3PO4/molar mass of K3PO4

=(3.53 g)/(212.27 g/mol)

= 0.0166 mol

calculate mol of AgNO3:

mol of AgNO3 = M(AgNO3)*V(AgNO3)

= 0.545 M * 0.0753 L

= 0.0410 mol

1 mol of K3PO4 needs 3 mol of AgNO3 to react completely

So, 0.0166 mol of K3PO4 needs, 3*0.0166 = 0.0498 mol of AgNO3

But we have only 0.0410 mol AgNO3

So, AgNO3 is limiting reagent

mol of Ag3PO4 formed = (1/3)*moles of AgNO3

= 1/3 * 0.410

= 0.137 moles

Molar mass of Ag3PO4,

MM = 3*MM(Ag) + 1*MM(P) + 4*MM(O)

= 3*107.9 + 1*30.97 + 4*16.0

= 418.67 g/mol

mass of Ag3PO4,

m = number of mol * molar mass

= 0.137 mol * 418.67 g/mol

= 57.4 g

This is theoretical yield

% yield = actual yield * 100 / theoretical yield

= 4.86*100/57.4

= 8.5 %

Answer: 8.5 %