[Kris44]

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]

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?

[Kris44]

1 mole of copper sulphate.

[Borek]

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?

[Kris44]

sorry 0.400 mol of copper sulphate?

[Sev]

Solution is 0.4 mol per dm³ (i.e. 0.4M).  To find number of moles you will need to use n=cv.

[Kris44]

ok so 0.4 x 0.02dm³ = 0.008mol dm³

[Borek]

Value is OK, but units... units... UNITS!!! are completely wrong.

[Kris44]

ok so just mols?

[Borek]

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

[Kris44]

1 Mol?

[Kris44]

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

[Borek]

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

[Kris44]

ok is it 0.008mol of copper oxide as ratio is 1:1?

Thankyou for your help.
Kris.

[Borek]

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.