anonymous
  • anonymous
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 %
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
i think 89.9% not sure i need to check
anonymous
  • anonymous
nope its not E 89.9% dont put that ok @danicap
anonymous
  • anonymous
lol okay

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Australopithecus
  • Australopithecus
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
Australopithecus
  • Australopithecus
A limiting reagent is the reagent used completely in a reaction
anonymous
  • anonymous
what do u think@ Australopithecus or @danicap
Australopithecus
  • Australopithecus
If you have problems with any of the steps let me know
anonymous
  • anonymous
me
Australopithecus
  • Australopithecus
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
anonymous
  • anonymous
okay, thank you.
anonymous
  • anonymous
ok thanks for the break down
anonymous
  • anonymous
i say -43.45 % @Australopithecus
anonymous
  • anonymous
@danicap what do you think
anonymous
  • anonymous
I've never been good at chemisty, it's my 2nd year on doing this lol. So honestly idk.
anonymous
  • anonymous
oh
anonymous
  • anonymous
it might be hard
anonymous
  • anonymous
I'm trying to do it on how he broke it down , but idk if i'm doing it right.
anonymous
  • anonymous
whats your answer
anonymous
  • anonymous
sorry
anonymous
  • anonymous
checking with my chemistry book
anonymous
  • anonymous
hold on , almost done.
anonymous
  • anonymous
so is it right
anonymous
  • anonymous
Yes, you were right. Thank you
anonymous
  • anonymous
awesome need more help
Australopithecus
  • Australopithecus
Well if you follow my method you should be correct unless you made a calculation error
anonymous
  • anonymous
all i have is a student book
anonymous
  • anonymous
because if i help some one i make sure that im right
anonymous
  • anonymous
The problem I have is I haven't been taught this. I'm teachng myself so it's a lot harder.
anonymous
  • anonymous
oh sorry
anonymous
  • anonymous
It's okay, thank you for the help :)
anonymous
  • anonymous
is it the same question?

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