Simple chemistry problem, just want to know if I did it right

Question

A precipitate of calcium phosphate is formed when aqueous solutions of sodium phosphate and excess calcium chloride are mixed. If 100.0 mL of a 1.5 M sodium phosphate solution is the limiting reactant, how many grams of calcium phosphate can be expected? (don't forget the aqueous solution as a product)

I got 46.53 grams, but can someone please try as well, so that i can see i got the right answer in steps please, thank you

Answer 1

The balanced equation is: 2Na3PO4 + 3CaCl2 = Ca3(PO4)2 + 6NaCl

Find the moles of sodium phosphate.  Moles = Molarity x Liters so moles = 1.5 M x .100 L = .15 moles.  Now look at the ratio of sodium phosphate to calcium phosphate: 2:1.  If .15 moles represents your 2, then .075 moles (1/2 of .15) would be your 1. The weight of Ca3(PO4)2 is:

Ca = 40 x 3 = 120 grams

P = 31 x 2 = 62 grams

O = 16 x 8 = 128 grams

Added together they total 310 grams / mole.  Multiply this by .075 moles = 23.25 grams expected.

Answer 2

I actually got an answer that is half of what you got. I got 23.25.. or maybe i made a mistake somewhere..

The balanced equation for the reaction is

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

2 moles           3moles      6 moles  1 mole

Using the limiting agent, 2 moles of Na3PO4 produces 1 mole of Ca3(PO4)2

Using the given concentration/molarity of 1.5M and the volume of 100.0ml, the number of moles is given by

n = CV, n = 1.5*100/1000 (moles) = 0.15 moles.

Therefore you know that 0.15 moles is used in the reaction.

So if 2 moles of Na3PO4 produces 1 mole of Ca3(PO4)2, what would 0.15 moles of Na3PO4 produce, that would be

0.15/2 = 0.075 moles of Na3PO4.

How much in grams is 0.075 moles of Na3PO4?

0.075 moles of Na3PO4 is 0.075*it's molar mass

that is 0.075*310 and that gives you 23.25 grams