What is the maximum number of grams of silver phosphate that can be produced from the reaction of 287g of silver chlorate with 52.8g of sodium phosphate? 3 ag(clo3) + na3(po4) —> 3 na(clo3) + ag3(po4)

The balanced equation for the above reaction is as follows;
3AgClO₃ + Na₃PO₄ ---> Ag₃PO₄ + 3NaClO₃
stoichiometry of AgClO₃ to Na₃PO₄ is 3:1
first we need to find which is the limiting reactant 

number of AgClO₃ moles - 287 g / 191.3 g/mol = 1.500 mol 
number of Na₃PO₄ moles - 52.8 g / 163.9 g/mol = 0.322 mol 
if AgClO₃ is the limiting reactant 
then 1.500 mol of AgClO₃ reacts with - 1.500/3 mol of Na₃PO₄
therefore number of Na₃PO₄ moles required = 0.50 mol of Na₃PO₄ are required but only 0.322 mol are present 
therefore Na₃PO₄ is the limiting reactant 
amount of product formed depends on amount of limiting reactant present 
stoichiometry of Na₃PO₄ to Ag₃PO₄ is 1:1
the number of Na₃PO₄ moles reacted - 0.322 mol
then number of Ag₃PO₄ moles formed - 0.322 mol
mass of Ag₃PO₄ formed - 0.322 mol x 418.6 g/mol = 134.8 g
mass of  Ag₃PO₄  produced is 134.8 g

The answer is c) pyrite