Question

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 %

Answer 1

i think 89.9% not sure i need to check

Answer 2

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

Answer 3

lol okay

Answer 4

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

Answer 5

A limiting reagent is the reagent used completely in a reaction

Answer 6

what do u think@ Australopithecus or
@danicap

Answer 7

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

Answer 8

me

Answer 9

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

Answer 10

okay, thank you.

Answer 11

ok thanks for the break down

Answer 12

i say -43.45 % @Australopithecus

Answer 13

@danicap what do you think

Answer 14

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

Answer 15

oh

Answer 16

it might be hard

Answer 17

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

Answer 18

whats your answer

Answer 19

sorry

Answer 20

checking with my chemistry book

Answer 21

hold on , almost done.

Answer 22

so is it right

Answer 23

Yes, you were right. Thank you

Answer 24

awesome need more help

Answer 25

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

Answer 26

all i have is a student book

Answer 27

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

Answer 28

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

Answer 29

oh sorry

Answer 30

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

Answer 31

is it the same question?