What is the theoretical yield of calcium phosphate?

  I'm going to assume that the other reactant is sodium phosphate(you did not indicate this) 
so i'm guessing that the reaction did this 
CaCL2 + Na3(PO4) > Ca3(PO4)2 + NaCl 
(since this reaction was done in water(aqueous solutions involved) the NaCl would be in the solution and thus the filtrate of the product so it won't influence the mass of the calcium phosphate product) 

so then you need to write the balanced equation for this reaction(lets assume you can do this) 
3 CaCL2 + 2 Na3(PO4) > Ca3(PO4)2 + 6 NaCl 

so in theoretical analysis 3 moles of calcium chloride is involved in the production of 1 mole of calcium phosphate 

in terms of molar mass equivalents 3 molesw of calcium chloride weigh = 3 x (40 + 71)g/mol = 333g 
1 mole of calcium phosphate = 310 g 
so if you only have 0.513 g of calcium chloride then this would calculate to get the mass of calcium phosphate  
mass = [0.513 / 333] x 310 g = 0.477grams  

your actual yield is what you massed when you performed the lab. hope that helped.

The gas would begin to expand, and eventually the can would burst or explode.