Can I get some help with this problem??

Question

You combined 86.0mL of 0.385M aqueous calcium nitrate solution with 80.0mL of 0.362M aqueous potassium phosphate solution to form a white precipitate. After you have filtered and dried the white precipitate, you have 2.014g of white solid. [The Ksp of the white precipitate is 1.73x10^-26].

a. Write a balanced chemical equation for the reaction that's taking place and identify the white

precipitate.

b. What is the theoretical yield of the white precipitate?

c. What is the percent yield? If the percent yield is not 100%, give reasons as to why it's not 100%.

Answer 1

a). 3Ca(NO₃)₂(aq) + 2K₃PO₄(aq) ==> Ca₃(PO₄)₂(s) + 6KNO₃(aq) ... balanced equation

The white precipitate is Ca₃(PO₄)₂ (calcium phosphate)

b). To find the theoretical yield we first need to find which reactant is limiting. An easy way to do this is to simply divide the moles of each reactant by the corresponding coefficient in the balanced equation and whichever value is less indicates the limiting reactant.

For Ca(NO3)2: 86.0 ml x 1 L / 1000 ml x 0.385 mol / L = 0.03311 mols Ca(NO₃)₂ (÷3->0.011)

For K3PO4: 80.0 ml x 1 L / 1000 ml x 0.363 mol / L = 0.0290 mols K3PO4 (÷ 2->0.015)

According to these calculations, Ca(NO3)2 will be limiting, but only by a slight margin.

Theoretical yield:

0.03311 mol Ca(NO3)2 x 1 mol Ca3(PO4)2 / 3 mol Ca(NO3)2 = 0.01104 mols Ca3(PO4)2

0.01104 mol Ca3(PO4)2 x 310.2 g / mol = 3.425 g Ca3(PO4)2

c). Percent yield:

actual / theoretical (x100%) = 2.014 g / 3.425 g (x100%) = 58.8% yield

There can be many reasons why % yield isn't 100%.

1. the reaction didn't go to completion
2. all of the precipitate wasn't allowed to settle and therefore wasn't all collected
3. loss of precipitate during transfer and poor laboratory technique