Answer to Question #273415 in General Chemistry for mia

Question

Consider the following balanced equation:

3NH4NO3 + Na3PO4 → (NH4)3PO4 + 3NaNO3

30 g NH4NO3 and 50 g Na3PO4 are allowed to react

-What is the maximum amount of each product that can be prepared?

Answer 1

Moles of Na3PO4= 50/164=0.3049moles

Moles of NH4NO3=30/80= 0.375 moles

NH4)3PO4 moles produced= 0.3049mole

NaNO3 moles produced= 0.3049×3= 0.9138