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