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### Topic: Solution/product calculation help  (Read 9364 times)

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#### Kris44

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##### Solution/product calculation help
« on: November 03, 2007, 07:49:12 PM »
Hello i am stuck on this problem and don't know how to approach it. The question is. A student set out to make some Copper(II) Oxide from Copper(II) Sulphate. He took 20cm3 of 0.400mol dm3 Copper(II) Sulphate solution, added an excess of Sodium Hydroxide to give a precipitate of Copper(II) Hydroxide and then boiled the mixture to convert the Copper(II) hydroxide into Copper(II) Oxide. What is the maximum mass of Copper(II) Oxide that he could obtain.

Formula+ CuSO4 + 2NaOH ----> Cu(OH)2 + Na2SO4

Cu(OH)2 ---->CuO + H2O

Any help really appreciated
regards,
Chris.

#### Borek

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##### Re: Solution/product calculation help
« Reply #1 on: November 03, 2007, 07:57:17 PM »
Do you know what reaction equation means when balanced? In this case it tells you, that for every mole of copper sulfate one mole of copper hydroxide is produced, and that further for every mole of copper hydroxide one moled copper oxide is produced. Or shorter - for every mole of copper sulfate one mole of copper oxide is produced.

How many moles of copper sulfate were used?
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#### Kris44

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##### Re: Solution/product calculation help
« Reply #2 on: November 04, 2007, 06:17:44 AM »
1 mole of copper sulphate.

#### Borek

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##### Re: Solution/product calculation help
« Reply #3 on: November 04, 2007, 07:13:57 AM »
1 mole of copper sulphate.

No:

He took 20cm3 of 0.400mol dm3 Copper(II) Sulphate solution

How much copper sulfate in this volume of the solution?
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#### Kris44

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##### Re: Solution/product calculation help
« Reply #4 on: November 04, 2007, 07:20:20 AM »
sorry 0.400 mol of copper sulphate?

#### Sev

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##### Re: Solution/product calculation help
« Reply #5 on: November 04, 2007, 07:56:08 AM »
Solution is 0.4 mol per dm3 (i.e. 0.4M).  To find number of moles you will need to use n=cv.

#### Kris44

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##### Re: Solution/product calculation help
« Reply #6 on: November 04, 2007, 08:44:01 AM »
ok so 0.4 x 0.02dm3 = 0.008mol dm3

#### Borek

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##### Re: Solution/product calculation help
« Reply #7 on: November 04, 2007, 09:30:45 AM »
Value is OK, but units... units... UNITS!!! are completely wrong.
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#### Kris44

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##### Re: Solution/product calculation help
« Reply #8 on: November 04, 2007, 09:39:26 AM »
ok so just mols?

#### Borek

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##### Re: Solution/product calculation help
« Reply #9 on: November 04, 2007, 10:27:20 AM »
Right - there were 0.008 moles of copper sulfate. How many moles of copper oxide were made? Remember - it is in the reaction equation.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations
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#### Kris44

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##### Re: Solution/product calculation help
« Reply #10 on: November 04, 2007, 11:02:19 AM »
1 Mol?

#### Kris44

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##### Re: Solution/product calculation help
« Reply #11 on: November 04, 2007, 11:09:45 AM »
ok so would i do now m=nM = (0.008 x 64+16)= 0.64g of copper oxide? as ratio is 1:1?

#### Borek

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##### Re: Solution/product calculation help
« Reply #12 on: November 04, 2007, 12:06:00 PM »
1 Mol?

No, but

ok so would i do now m=nM = (0.008 x 64+16)= 0.64g of copper oxide? as ratio is 1:1?

yes
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#### Kris44

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##### Re: Solution/product calculation help
« Reply #13 on: November 04, 2007, 01:06:16 PM »
ok is it 0.008mol of copper oxide as ratio is 1:1?

Kris.

#### Borek

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##### Re: Solution/product calculation help
« Reply #14 on: November 04, 2007, 02:09:49 PM »
Take a look at both reactions - in each case 1 mole of substance containing copper produces one mole of substance containing copper. Thus if you start with 1 mole of copper sulfate - you end with one mole of copper oxide. If you start with 3 moles of copper sulfate - you end with 3 moles of copper oxide. If you start with half mole of copper sulfate - you end with half mole of copper oxide - and so on. You start with 0.008 mole of sulfate - and you end with 0.008 mole of oxide.

That's not always the case (this 1:1 ratio) - but ratio will be always defined by the reaction equation.
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