Question

If you mix a 25.0mL sample of a 1.2M potassium phosphate solution with 30.00mL of a 1.10M barium nitrate solution, a precipitation reaction occurs. You collect 3.31g of solid dry precipitate. What is the % yield in your reaction?

K3PO4+Ba(NO3)2---> KNO3+Ba3(PO4)2

Answer 1

volume of K3PO4, V = 25.0 mL

= 2.5*10^-2 L

we have below equation to be used:

number of mol in K3PO4,

n = Molarity * Volume

= 1.2*0.025

= 3*10^-2 mol

volume of Ba(NO3)2, V = 30.0 mL

= 3*10^-2 L

we have below equation to be used:

number of mol in Ba(NO3)2,

n = Molarity * Volume

= 1.1*0.03

= 3.3*10^-2 mol

we have the Balanced chemical equation as:

2 K3PO4 + 3 Ba(NO3)2 ---> Ba3(PO4)2 + 6 KNO3

2 mol of K3PO4 reacts with 3 mol of Ba(NO3)2

for 3*10^-2 mol of K3PO4, 4.5*10^-2 mol of Ba(NO3)2 is required

But we have 3.3*10^-2 mol of Ba(NO3)2

so, Ba(NO3)2 is limiting reagent

we will use Ba(NO3)2 in further calculation

Molar mass of Ba3(PO4)2 = 3*MM(Ba) + 2*MM(P) + 8*MM(O)

= 3*137.3 + 2*30.97 + 8*16.0

= 601.84 g/mol

From balanced chemical reaction, we see that

when 3 mol of Ba(NO3)2 reacts, 1 mol of Ba3(PO4)2 is formed

mol of Ba3(PO4)2 formed = (1/3)* moles of Ba(NO3)2

= (1/3)*3.3*10^-2

= 1.1*10^-2 mol

we have below equation to be used:

mass of Ba3(PO4)2 = number of mol * molar mass

= 1.1*10^-2*6.018*10^2

= 6.62 g

% yield = actual mass*100/theoretical mass

= 3.31*100/6.62

= 50.0 %

Answer: 50.0 %