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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 %
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think 89.9% not sure i need to check

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nope its not E 89.9% dont put that ok @danicap

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.01. 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
 one year ago
Best ResponseYou've already chosen the best response.0A limiting reagent is the reagent used completely in a reaction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what do u think@ Australopithecus or @danicap

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0If you have problems with any of the steps let me know

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0Also 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
 one year ago
Best ResponseYou've already chosen the best response.0ok thanks for the break down

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i say 43.45 % @Australopithecus

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@danicap what do you think

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've never been good at chemisty, it's my 2nd year on doing this lol. So honestly idk.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm trying to do it on how he broke it down , but idk if i'm doing it right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0checking with my chemistry book

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hold on , almost done.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, you were right. Thank you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0awesome need more help

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.0Well if you follow my method you should be correct unless you made a calculation error

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0all i have is a student book

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because if i help some one i make sure that im right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The problem I have is I haven't been taught this. I'm teachng myself so it's a lot harder.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's okay, thank you for the help :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it the same question?
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