Question

Aqueous potassium phosphate is added to 545 mL of a solution containing calcium chloride to precipitate all of the Ca2+ ions as the insoluble phosphate (310.2 g/mol). If 4.51 g calcium phosphate is produced, what is the Ca2+ concentration in the calcium chloride solution in g/L?

Answer 1

2K3PO4 + 3CaCl2 ---------- Ca3(PO4)2 + 6KCl

According to the balanced equation we have that 3 mol of CaCL2 produces 1 mol of Ca3(PO4). Now we have to change 4,51g to mol with the molecular weigth.

1 mol of Ca3(PO4) -----weights----- 310,2g

X --------------------------------------------- 4,51g

X= 0,0145mol of Ca3(PO4)

3 mol of CaCl2 ----- produces--------- 1 mol Ca3(PO4)

X ----------------------------------------------- 0,0145 mol Ca3(PO4)

X= 0,0436 mol of CaCl2

Now this amount we have to change it to grams with the molecular weight, 110,98g/mol

1 mol of CaCl2 -----weights------- 110,98g

0,0436 mol ---------------------------- X

X= 4,84g of CaCl2.

Now they are asking us to express the concentration on g/L, we already have the grams tht we have calculate, now we have to change the mL to L.

4,84g CaCl2 ----------are in --------- 0,545L

X ------------------------------------------- 1L

X= 8,88g/L

This means that you will have 8,88g of CaCl2 for each liter of solution that you have.