Answer to Question #118551 in General Chemistry for nani

Question

What is the maximum number of grams of silver phosphate that can be produced from the reaction of 287 g of silver chlorate with 52.8 g of sodium phosphate? 3 Ag(ClO3) + Na3(PO4)

Answer 1

3AgClO₃ + Na₃PO₄ → Ag₃PO₄ + 3NaClO₃

M(AgClO₃) = 191.32 g/mol;

M(Na₃PO₄) = 163.94 g/mol;

M(Ag₃PO₄) = 418.58 g/mol;

n=m/M;

n(AgClO₃) = 287 g / (191.32 g/mol) = 1.5 moles;

n(Na₃PO₄) = 52.8/ (163.94 g/mol) = 0.32 moles;

Na₃PO₄ is limiting reactant;

According to the reaction:

n(Ag₃PO₄) = n(Na₃PO₄) = 0.32 moles;

m = n×M;

m(Ag₃PO₄) = 0.32 moles × 418.58 g/mol = 133.94 g

Answer: 133.94 g