The yellow compound silver(I) phosphate may be formed via the following reaction.
Na3PO4 + 3 AgNO3 = Ag3PO4 + 3 NaNO3
If you add an aqueous solution that contains 0.40 moles of Na3PO4 and an aqueous solution that contains 0.66 moles of AgNO3 to a reaction container, when the reaction is complete
A. 0.22 moles of NaNO3 will be made
B. 0.18 moles of Na3PO4 will be left over and all AgNO3 will be used
C. some AgNO3 will be left-over but all Na3PO4 will be reacted
D. no Na3PO4 and no AgNO3 will be left over
Is it B?
We don't know which is the limiting reagent (LR) so I'll do two problems; one assuming Na3PO4 is the LR and the other assuming AgNO3 is the LR.
First AgNO3:
.................Na3PO4 + 3 AgNO3 = Ag3PO4 + 3 NaNO3
Initial...........0.40..............0.00.............0................0
add...................................0.66................................
change.......-0.66/3 =
....................-0.22.............-0.66.........+0.22...........0.66
equilibrium...0.18..................0..............0.22...........0.66
I guessed right to begin. Yes, B is correct. But let's do the other one to make sure.
.........................Na3PO4 + 3 AgNO3 = Ag3PO4 + 3 NaNO3
initial......................0................0.66............0.....................0
add......................0.40............................................................
change...............-0.40............ -3*0.4 =
...................................................-1.20...............but we don't have 1.20; i.e.,we have only 0.66 so this is not possible.
I go with B.