# Percentage yield HELP!!!

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Wolfram Alpha

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#1

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?

a) If 15 grams of copper (II) chloride react with 20. grams of sodium nitrate,how much sodium chloride can be formed?

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Dylann

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#2

(Original post by

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?

**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?

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)

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Wolfram Alpha

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#3

(Original post by

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)

**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)

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Dylann

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#4

(Original post by

Thank you ;p but how to find the mole?

**Wolfram Alpha**)Thank you ;p but how to find the mole?

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!

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Fallen99

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#5

**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

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FireFreak

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#6

(Original post by

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

**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

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FireFreak

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#7

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#7

**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.

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Fallen99

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#8

(Original post by

This is not the right method or the right answer.

**FireFreak**)This is not the right method or the right answer.

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Fallen99

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#9

(Original post by

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

**FireFreak**)There is not a 1:1 ratio. The equation is not balanced.

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FireFreak

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#10

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.

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.

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Dylann

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#11

(Original post by

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

**FireFreak**)There is not a 1:1 ratio. The equation is not balanced.

So you have to write the balanced equation first:

CuCl2 + 2NaNO3 Cu(NO3)2 + 2NaCl

(Original post by

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...

**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...

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...

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#12

(Original post by

This is not the right method or the right answer.

**FireFreak**)This is not the right method or the right answer.

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

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#13

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

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#14

(Original post by

Lol didn't even notice that

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...

**Dylann**)Lol didn't even notice that

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...

And y cant u add the masses together??

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#16

(Original post by

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

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

**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.

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#17

**Dylann**)

Lol didn't even notice that

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...

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#18

(Original post by

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

**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...

Thanks btw

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#19

(Original post by

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.

**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.

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#20

(Original post by

Correct !

**FireFreak**)Correct !

(Original post by

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

**Fallen99**)Doesn't the coefficient tells u that there are two moles of Na in this equation??

2x + 5y ---> 3z + 4w

This means that every 2 moles of x react with 5 moles of y producing 3 moles of z and 4 moles of w

It doesn't tell you how many moles of each substance you actually have...just the RATIO in which they react...very important you don't get confused...

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