## 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

i think 89.9% not sure i need to check

2. anonymous

nope its not E 89.9% dont put that ok @danicap

3. anonymous

lol okay

4. 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

5. Australopithecus

A limiting reagent is the reagent used completely in a reaction

6. anonymous

what do u think@ Australopithecus or @danicap

7. Australopithecus

If you have problems with any of the steps let me know

8. anonymous

me

9. 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

10. anonymous

okay, thank you.

11. anonymous

ok thanks for the break down

12. anonymous

i say -43.45 % @Australopithecus

13. anonymous

@danicap what do you think

14. anonymous

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

15. anonymous

oh

16. anonymous

it might be hard

17. anonymous

I'm trying to do it on how he broke it down , but idk if i'm doing it right.

18. anonymous

19. anonymous

sorry

20. anonymous

checking with my chemistry book

21. anonymous

hold on , almost done.

22. anonymous

so is it right

23. anonymous

Yes, you were right. Thank you

24. anonymous

awesome need more help

25. Australopithecus

Well if you follow my method you should be correct unless you made a calculation error

26. anonymous

all i have is a student book

27. anonymous

because if i help some one i make sure that im right

28. anonymous

The problem I have is I haven't been taught this. I'm teachng myself so it's a lot harder.

29. anonymous

oh sorry

30. anonymous

It's okay, thank you for the help :)

31. anonymous

is it the same question?