Hey bob- Here's what I got so far:
3AgNO3 + Na3PO4 ---> Ag3PO4 + 3NaNO3
Silver nitrate and sodium phosphate are reacted in equal amounts of 200. g each. How many grams of silver phosphate are produced?
1. What is the limiting?
2. How much silver phosphate is produced?
3. How much is in excess?
200gAGNO3/170= 1.18 AGNO3
200gNA3PO4/164= 1.22 Ag3PO4
Not sure of the next step....here's what I got:
419 AG3PO4 x 3 = 1257 AG3PO4
Is AG3PO4 the limiting reagent??
Can you check my answer from your last tips:
1. Ok I think I understand. Still a little unsure tho where we're getting the 3 times the silver nitrate from for silver nitrate to be limiting reagent.
2. 1.18/3=.39 x 419 Ag3PO4 = 163g Ag3PO4 produced??
200gAGNO3/170= 1.18 AGNO3
200gNA3PO4/164= 1.22 Ag3PO4
3AgNO3 + Na3PO4 ---> Ag3PO4 + 3NaNO3
One needs three time the silver nitrate, you dont have near that, so silver nitrate is the limiting reageant. So you will consume all you have 1.18 moles, and you will get 1.18/3 moles of silver phosphate.
On the sodium phosphate, you will consume 1.18/3 moles of it, you had 1.22 to start, so the left over is the difference, and you cna convert that to grams
3. 200g Na3PO4 - 163 = 36g Ag3PO4??
The three times is because your balanced equation says you need three silver nitrate molecules for every sodium phosphate molecule.
You do not have that much silver nitrate so it limits the reaction.
Gotcha, thanks. Can you one of you guys see if I got 2 & 3 correct from the tips bob was giving me.
so you use 1.18 mols of silver nitrate
that needs 1.18/3 = .393 mols of sodium phosphate. For every mol of sodium phosphate you get a mol of silver phosphate so you get ,393 mols of Silver Phosphate
you have 1.22 mols of sodium phosphate
so you will have 1.22 - .39 = .83 mols of sodium phosphate left over.
I will leave you to convert mols to grams :)
I only do mols.
3.) Cool that makes sense. So I believe this is how you convert mole to gram:
.83 x 164g/mol = 136g Na3PO4 left over