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anonymous

  • one year ago

PLEASE HELP!! When 15 grams of copper (II) chloride (CuCl2) reacts with 20 grams of sodium nitrate (NaNO3), 11.3 grams of sodium chloride (NaCl) is obtained. What is the percent yield of this reaction? CuCl2 + 2NaNO3 Cu(NO3)2 + 2NaCl -11.3 % -13.1 % -26.2 % -43.45 % -86.9 %

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  1. anonymous
    • one year ago
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    i think 89.9% not sure i need to check

  2. anonymous
    • one year ago
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    nope its not E 89.9% dont put that ok @danicap

  3. anonymous
    • one year ago
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    lol okay

  4. Australopithecus
    • one year ago
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    1. First step you need to figure out what your limiting reagent the limiting reagent is convert copper (II) chloride and sodium nitrate to moles mole = grams/molecular mass 2. using your reaction figure out which reactant gives you the smallest amount of moles of NaCl, that is your limiting reactant. Also doing this you now have your theoretical yield (the maximum amount of product (NaCl) you could obtain from the reaction) 3. Find moles of NaCl that is your actual yield 4. Plug theoretical yield and actual yield into the formula: \[Percent\ Theoretical\ Yield = (\frac{Actual\ Yield}{Theoretical\ Yield})*100\] This will give you percent yield of NaCl

  5. Australopithecus
    • one year ago
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    A limiting reagent is the reagent used completely in a reaction

  6. anonymous
    • one year ago
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    what do u think@ Australopithecus or @danicap

  7. Australopithecus
    • one year ago
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    If you have problems with any of the steps let me know

  8. anonymous
    • one year ago
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    me

  9. Australopithecus
    • one year ago
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    Also you can just say, If I had 1 mole of CuCl2 and 1 mole of NaNO3 which would produce the least amount of NaCl, that will give you the limiting reagent of your reaction

  10. anonymous
    • one year ago
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    okay, thank you.

  11. anonymous
    • one year ago
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    ok thanks for the break down

  12. anonymous
    • one year ago
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    i say -43.45 % @Australopithecus

  13. anonymous
    • one year ago
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    @danicap what do you think

  14. anonymous
    • one year ago
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    I've never been good at chemisty, it's my 2nd year on doing this lol. So honestly idk.

  15. anonymous
    • one year ago
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    oh

  16. anonymous
    • one year ago
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    it might be hard

  17. anonymous
    • one year ago
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    I'm trying to do it on how he broke it down , but idk if i'm doing it right.

  18. anonymous
    • one year ago
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    whats your answer

  19. anonymous
    • one year ago
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    sorry

  20. anonymous
    • one year ago
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    checking with my chemistry book

  21. anonymous
    • one year ago
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    hold on , almost done.

  22. anonymous
    • one year ago
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    so is it right

  23. anonymous
    • one year ago
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    Yes, you were right. Thank you

  24. anonymous
    • one year ago
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    awesome need more help

  25. Australopithecus
    • one year ago
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    Well if you follow my method you should be correct unless you made a calculation error

  26. anonymous
    • one year ago
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    all i have is a student book

  27. anonymous
    • one year ago
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    because if i help some one i make sure that im right

  28. anonymous
    • one year ago
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    The problem I have is I haven't been taught this. I'm teachng myself so it's a lot harder.

  29. anonymous
    • one year ago
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    oh sorry

  30. anonymous
    • one year ago
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    It's okay, thank you for the help :)

  31. anonymous
    • one year ago
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    is it the same question?

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