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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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
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i think 89.9% not sure i need to check
nope its not E 89.9% dont put that ok @danicap
lol okay

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Other answers:

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

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