How many grams of calcium phosphate can be produced when 89.3 grams of calcium chloride reacts with excess sodium phosphate?

Molecular weights: 
CaCl2 = 40 + 35.5*2 = 111 g/mol 
Ca3(PO4)2 = 40*3 + (31+16*4)*2 = 310 g/mol 
89.3 g CaCl2 = 89.3/111 = 0.8045 mol CaCl2 

Balanced equation:
          3CaCl2 + 2Na3PO4 > 6NaCl + Ca3(PO4)2 

3 moles CaCl2 produce 1 mol Ca3(PO4)2 
Therefore 0.8045 mol CaCl2 produces 1/3 * 0.8045 
= 0.2682 mol Ca3(PO4)2 
= 0.2682 mol * 310 g/mol 
= 83.1 g Ca3(PO4)2 

Ans: 83.1 g 

Answer is to avoid experimental error