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?
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
