Question #273415
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?
Expert's answer
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
