Calculating the mass of carbon needed to reduce 15.9g of Copper(II) Oxide to Copper

  • Thread starter ViralRiver
  • Start date
  • #1
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Homework Statement



Calculate what mass of carbon you would need to reduce 15.9g of copper(II) oxide to copper by the reaction:

CuO(s) + C(s) [tex]\rightarrow[/tex] Cu(s) + CO(s)

Homework Equations



n = m / ar
(number of moles = mass / relative atomic mass)

If you know any more, it would be really useful for me though ;) .

The Attempt at a Solution



Ar(Cu) = 63.5
Ar(O) = 16.0
Ar(C) = 12.0

CuO = 1 / (63.5 + 16.0) = 79.5g
C = 1 / (12.0) = 12.0g

79.5 / 79.5 * 15.9 = 15.9g of CuO
12.0 / 79.5 * 15.9 = 2.4g of C

Basically, I divided the mass of CuO by the mass of CuO and multiplied by 15.9 to get 15.9g (as in the question). Then, I applied the same equation to C by dividing the mass of C by 79.5 and then multiplying by 15.9 to get 2.4g, which is my answer.

Is this correct?
 

Answers and Replies

  • #2
867
0


Your answer is correct (don't quite understand all your steps, but your figures are right). How many significant figures do you need in your answer?

I would have done it like this:

[tex]15.9g ~CuO \left(\frac{1 ~mol ~CuO}{79.5g ~CuO}\right)\left(\frac{1 ~mol~ C}{1 ~mol ~CuO}\right)\left(\frac{12.01g ~C}{1 ~mol ~C}\right) = grams ~of ~C[/tex]
 
Last edited:
  • #3
Borek
Mentor
28,564
3,012


I would have done it like this:

[tex]15.9g ~CuO \left(\frac{1 ~mol ~CuO}{79.5g ~CuO}\right)\left(\frac{1 ~mol~ C}{1 ~mol ~CuO}\right)\left(\frac{12.01g ~C}{1 ~mol ~C}\right) = grams ~of ~C[/tex]
Please note, that dimensional analysis is not universally taught and in many places of the world stoichiometry is done with ratios.

Which doesn't mean you are wrong, just the OP can be completely not aware of the method you propose.
 
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