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??

15 answers

  1. 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

  2. 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??

  3. 3. 200g Na3PO4 - 163 = 36g Ag3PO4??

  4. The three times is because your balanced equation says you need three silver nitrate molecules for every sodium phosphate molecule.

  5. You do not have that much silver nitrate so it limits the reaction.

  6. Gotcha, thanks. Can you one of you guys see if I got 2 & 3 correct from the tips bob was giving me.

  7. 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.

  8. I will leave you to convert mols to grams :)

  9. I only do mols.

  10. 3.) Cool that makes sense. So I believe this is how you convert mole to gram:

    .83 x 164g/mol = 136g Na3PO4 left over

  11. I assume your 164 g/mol is correct. I know O is 16 and Na is 23 but do not remember P . Not about to Google.

  12. Phosphorus is 31g.

  13. Ya and that equals 164. Thank you a ton for the help I've stomped on this whole problem for like a half an hour and I got lots more to do. This was my 2nd limiting reagent problem I've ever done so I'm still learning how to handle these.

  14. Would you have another 5 minutes to help me one more problem I'm stumped on???

  15. I'll post it as a new problem. Its a Stoichemetric Problem that I'm completely stumped on. I think I have part of correct which i'd be happy to see if its correct so far or not.

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