Percentage yield HELP!!!

Original post by Wolfram Alpha

CuCl2 + NaNO3  Cu(NO3)2 + NaCl
a) If 15 grams of copper (II) chloride react with 20. grams of sodium nitrate,how much sodium chloride can be formed?

Work out the moles of each of the reactants...see which one will be used up first..the reaction stops there...work out how many moles reacted...~~moles reacted = moles produced because of 1:1 ratio..~~ Write a balanced equation first and then work out the molar ratio...convert moles back to mass using Mr...

edit: And this is your theoretical maximum yield. You need some more information to work out the percentage yield (i.e. how much NaCl was actually formed)

(edited 9 years ago)

Reply 2

Wolfram Alpha

OP

20

Original post by Dylann

Work out the moles of each of the reactants...see which one will be used up first..the reaction stops there...work out how many moles reacted...moles reacted = moles produced because of 1:1 ratio...convert moles back to mass using Mr...

edit: And this is your theoretical maximum yield. You need some more information to work out the percentage yield (i.e. how much NaCl was actually formed)

Thank you ;p but how to find the mole?

Reply 3

Original post by Wolfram Alpha

Thank you ;p but how to find the mole?

mol=Mass in grams/Mr

Mr = Relative molecular mass (sum of all relative atomic masses) i.e. Mr of CuCl2 = 63.5 + (2x35.5) =134.5

Quite a fundamental formula that you should certainly know!

Reply 4

Original post by Wolfram Alpha

CuCl2 + NaNO3  Cu(NO3)2 + NaCl
a) If 15 grams of copper (II) chloride react with 20. grams of sodium nitrate,how much sodium chloride can be formed?

Use this formula= mass= moles*relative formula mass
So first work out the relative formula for each of them. Lets start with CuCl2
Cu= 63.5
Cl= 35.5*2= 71
Than it is NaNO3
Na= 23
Nitrogen= 14
Oxygen =16*3 = 48
Add the relative formula mass of CuCl2+NaNO3 together= 219.5
And you have the total mass of 15+20=35 grams
Use this to work out moles which is mass/ relative formula mass
Moles= 35/219.5= 0.159453303
Now work out the relative mass for NaCl
Na= 23
Cl= 35.5
Add them together= 58.5
For this you have to find mass which is moles* relative mass
0.159453303*58.5= 9.328018226
So 9.328018226 NaCl can be formed

Is this right I just did it so I don’t know

Hope this helps you :smile:

Reply 5

Original post by Fallen99

Use this formula= mass= moles*relative formula mass
So first work out the relative formula for each of them. Lets start with CuCl2
Cu= 63.5
Cl= 35.5*2= 71
Than it is NaNO3
Na= 23
Nitrogen= 14
Oxygen =16*3 = 48
Add the relative formula mass of CuCl2+NaNO3 together= 219.5
And you have the total mass of 15+20=35 grams
Use this to work out moles which is mass/ relative formula mass
Moles= 35/219.5= 0.159453303
Now work out the relative mass for NaCl
Na= 23
Cl= 35.5
Add them together= 58.5
For this you have to find mass which is moles* relative mass
0.159453303*58.5= 9.328018226
So 9.328018226 NaCl can be formed

Is this right I just did it so I don’t know

Hope this helps you :smile:

This is not the right method or the right answer.

Reply 6

Original post by Dylann

Work out the moles of each of the reactants...see which one will be used up first..the reaction stops there...work out how many moles reacted...moles reacted = moles produced because of 1:1 ratio...convert moles back to mass using Mr...

edit: And this is your theoretical maximum yield. You need some more information to work out the percentage yield (i.e. how much NaCl was actually formed)

There is not a 1:1 ratio. The equation is not balanced.

Reply 7

Original post by FireFreak

This is not the right method or the right answer.

Ok than u can help me too

Reply 8

Original post by FireFreak

There is not a 1:1 ratio. The equation is not balanced.

I just noticed it too

Reply 9

1) Work out the balanced equation.
2) work out moles of each of your reactants. (Do not combine them....)
3) From your balanced equation work out the ratio of CuCl2 : NaNO3 (Should be 1:2)
4) 1 mole of Cucl2 will react with 2 moles of NaNO3, so times your mole of Cucl2 by two to give them amount of moles that will react with NaNO3 since there is a 1:2 ratio... Now the number you multiplied by two is the moles of Nacl produced since it will have a 2:2 ratio with Nano3 which is basically a 1:1.
5) Times the moles of NaCl produced (which is same as mole of NaNO3 present) by the mr to give the mass of NaCl produced.

Hope this helps.

Reply 10

Original post by FireFreak

There is not a 1:1 ratio. The equation is not balanced.

Lol didn't even notice that :rolleyes:

So you have to write the balanced equation first:

CuCl2 + 2NaNO3  Cu(NO3)2 + 2NaCl

Original post by Fallen99

Use this formula= mass= moles*relative formula mass
So first work out the relative formula for each of them. Lets start with CuCl2
Cu= 63.5
Cl= 35.5*2= 71
Than it is NaNO3
Na= 23...

You cannot add the total masses up and divide by total Mr...you must work out the moles individually as there will probably be some reactant left over...

moles CuCl2 = 15/134.5 =0.111...
moles NaNO3 = 20/85 = 0.235...

For every one mole of CuCl2, two moles of NaNO3 react...so if 0.111...moles of CuCl2 react you need (0.111..x 2 = 0.223...moles) of NaNO3...

So all the CuCl2 is used up with a little bit of NaNO3 left over. The moles of NaNO3 that I just worked out that react are 0.223...moles

From the balanced equation you can see that moles of NaNO3 = moles of NaCl...so 0.223... moles of NaCl formed...0.223 x 58.5 = 13.05 grams...

(edited 9 years ago)

Reply 11

Original post by FireFreak

This is not the right method or the right answer.

Sorry but the method is right I just did it again with the balanced equation and the answer is
Use this formula= mass= moles*relative formula mass
So first work out the relative formula for each of them. Lets start with CuCl2
Cu= 63.5
Cl= 35.5*2= 71
Than it is 2NaNO3
Na= 23*2=46
Nitrogen= 14
Oxygen =16*3 = 48
Add the relative formula mass of CuCl2+NaNO3 together= 242.5
And you have the total mass of 15+20=35 grams
Use this to work out moles which is mass/ relative formula mass
Moles= 35/242.5= 0.1443298962
Now work out the relative mass for 2NaCl
Na= 23*2= 46
Cl= 35.5*2= 71
Add them together= 117
For this you have to find mass which is moles* relative mass
0.1443298962*117= 13.37659786
So 13.37659786 Nacl can be formed

Reply 12

You should end up with an answer of 13.1g depending on if you have rounded your numbers or not.

Reply 13

Original post by Dylann

Lol didn't even notice that :rolleyes:

So you have to write the balanced equation first:

CuCl2 + 2NaNO3  Cu(NO3)2 + 2NaCl

You cannot add the total masses up and divide by total Mr...you must work out the moles individually as there will probably be some reactant left over...

moles CuCl2 = 15/134.5 =0.111...
moles NaNO3 = 20/85 = 0.235...

For every one mole of CuCl2, two moles of NaNO3 react...so if 0.111...moles of CuCl2 react you need (0.111..x 2 = 0.223...moles) of NaNO3...

So all the CuCl2 is used up with a little bit of NaNO3 left over. The moles of NaNO3 that I just worked out that react are 0.223...moles

From the balanced equation you can see that moles of NaNO3 = moles of NaCl...so 0.223... moles of NaCl formed...0.223 x 58.5 = 13.05 grams...

My answer is 13.37659786
And y cant u add the masses together??

Reply 14

Original post by Fallen99

My answer is 13.37659786
And y cant u add the masses together??

You can't add the masses together because there are different amounts of moles reacting...one reactant will be used up before the other...

Reply 15

Original post by Fallen99

Sorry but the method is right I just did it again with the balanced equation and the answer is
Use this formula= mass= moles*relative formula mass
So first work out the relative formula for each of them. Lets start with CuCl2
Cu= 63.5
Cl= 35.5*2= 71
Than it is 2NaNO3
Na= 23*2=46
Nitrogen= 14
Oxygen =16*3 = 48
Add the relative formula mass of CuCl2+NaNO3 together= 242.5
And you have the total mass of 15+20=35 grams
Use this to work out moles which is mass/ relative formula mass
Moles= 35/242.5= 0.1443298962
Now work out the relative mass for 2NaCl
Na= 23*2= 46
Cl= 35.5*2= 71
Add them together= 117
For this you have to find mass which is moles* relative mass
0.1443298962*117= 13.37659786
So 13.37659786 Nacl can be formed

Do not combine your masses or your Mr values. And also do not multiply the mr of sodium (highlighted in bold) by two, although the ratio is 1:2 this just tells you the amount reacting, so dont multiply by the coefficient.

Reply 16

Original post by Dylann

Lol didn't even notice that :rolleyes:

So you have to write the balanced equation first:

CuCl2 + 2NaNO3  Cu(NO3)2 + 2NaCl

You cannot add the total masses up and divide by total Mr...you must work out the moles individually as there will probably be some reactant left over...

moles CuCl2 = 15/134.5 =0.111...
moles NaNO3 = 20/85 = 0.235...

For every one mole of CuCl2, two moles of NaNO3 react...so if 0.111...moles of CuCl2 react you need (0.111..x 2 = 0.223...moles) of NaNO3...

So all the CuCl2 is used up with a little bit of NaNO3 left over. The moles of NaNO3 that I just worked out that react are 0.223...moles

From the balanced equation you can see that moles of NaNO3 = moles of NaCl...so 0.223... moles of NaCl formed...0.223 x 58.5 = 13.05 grams...

Correct

!

Reply 17

Original post by Dylann

You can't add the masses together because there are different amounts of moles reacting...one reactant will be used up before the other...

Okk I get it I never did a question like this before so it was something new
Thanks btw :smile:

Reply 18

Original post by FireFreak

Do not combine your masses or your Mr values. And also do not multiply the mr of sodium (highlighted in bold) by two, although the ratio is 1:2 this just tells you the amount reacting, so dont multiply by the coefficient.

Doesn't the coefficient tells u that there are two moles of Na in this equation??

Reply 19