I'm doing the reaction 2 CuO + C = 2 Cu + CO2. How many grams of CuO and C should I mix together to make sure there's none left, and all I have left
is pure copper powder? 2 grams of CuO and 1 gram of carbon?
I made the carbon with sugar and sulfuric acid BTWelementcollector1 - 5-6-2013 at 11:39
http://theodoregray.com/PeriodicTable/MSP/BalanceReactions for the easy method.
For the harder method, you'll need the molar mass of carbon, the molar mass of CuO, and the gram weight of one of these that you intend to use (say,
50). Using the molar masses, convert your grams of A to moles of A (where A can be either one). Then, using the balanced equation, convert moles of A
to moles of B (ex: 5 moles CuO for every 3 moles carbon - whatever the balanced numbers are). Finally, convert moles of B to grams of B. Make sure to
write these down as fractions (ex. 50g CuO x 1 mole CuO/999g CuO), making sure each unit *except* your final unit (g of B) is on both the top and the
bottom of any fraction at least once to properly cancel out the units.
And that's stoichiometry in a nutshell. This ensures that, to a reasonable degree, both reactants are used up equally, and close to no unused
reactants will be left mixed with the products.
On to the reaction itself:
You do know that reactions with carbon typically take place at 1000 Celsius in blast furnaces designed to withstand molten metal? If your reaction
works at all, you won't have any copper powder: You'll have a lump of copper metal.
To get a decent grade of copper powder, dissolve your CuO in any acid (even vinegar will work, and you can do the stoichiometry for this too if you
wish). Then, add a piece of aluminum foil and wait (probably overnight). What was formerly a blue solution of a soluble copper salt should be clear,
the Al should be gone or severly degraded (stoichiometry here too, why not? You don't want aluminum in your copper), and copper powder should be left
on the bottom of the container. You can clean this with almost any acid, except for HNO3, as copper metal is resistant (though not quite immune) to
chemical attack by most major acids.Cou - 5-6-2013 at 11:58
I'm doing the reaction 2 CuO + C = 2 Cu + CO2. How many grams of CuO and C should I mix together to make sure there's none left, and all I have left
is pure copper powder? 2 grams of CuO and 1 gram of carbon?
There are two answers to this question.
The first answer is the theoretical one, where everything works perfectly.
In this world you need 2 moles of CuO plus one mole of Carbon.
From the Periodic table, we get the following weights per mole:
Cu = 63.5
O = 16
Adding those up, a mole of CuO weighs 79.5 grams. We said we needed 2 moles of this, so 79.5 * 2 = 159.
Next we look up the carbon.
C = 12
So, we need (63.5 + 16) * 2 = 159 grams of CuO plus 12 grams of C.
Physicists refer to this wonderland where everything works perfectly as "A frictionless vacuum". In real life they don't exist. The second answer to
your question is the "real world" answer.
In the real world you'll use a gross excess of Carbon to reduce the CuO, and the reaction needs to be in a crucible that seals out atmospheric Oxygen.
Three to five times the amount required by the theoretical answer is not unreasonable. The excess Carbon is required to consume the atmospheric
Oxygen in the crucible plus any extra oxygen that sneaks into the crucible.
When your reaction is done, you'll have a lump of copper metal and unreacted carbon. The excess carbon can be converted to CO and CO2 by
heating while exposed to air.
There are two key concepts here:
1. The atomic weight from the periodic table is the weight in grams of one mole of an element. Add these weights up to get the perfect world amounts
of reactants you need for a reaction.
2. The real-world amounts can vary wildly from the the theoretical amounts due to side reactions and Murphy's law. This is what makes chemistry fun! Metacelsus - 12-6-2013 at 14:25
I did the carbothermal reduction a few months back with basic copper carbonate (made by mixing solutions of CuSO4 and Na2CO3) and charcoal. I had to
heat the crucible (covered, but not airtight, so as to let the evolved gas escape) until the bottom started to glow red, and I got my copper in the
form of fine powder. I used the stoichiometric amount of reactants, and I got significant unreacted basic copper carbonate which decomposed to copper
oxide. I was able to dissolve the copper oxide in hydrochloric acid, leaving fine copper powder. My yield was slightly less than 50%.
Carbon is cheap. Use an excess.watson.fawkes - 12-6-2013 at 16:25
It's basically impossible to do this reaction without also
doing the reaction CuO + C → Cu + CO, and furthermore, you'll have very little control over the ratio between these two reactions.plante1999 - 12-6-2013 at 16:39
boudouard reaction logic, every metallurgy chemist knows that.blogfast25 - 13-6-2013 at 03:09
Nicely explained by Elisabeth there.
There exist of course reactions that proceed perfectly stoichiometrically. See e.g. acid/base neutralisations in acidometry. An excess of one of the
reagents is thus not always a requirement.
For more fun, heat the CuO/excess carbon mixture to over the MP of Cu. The excess carbon then ends up floating on top of the molten Cu, slowly glowing
away.
This also works with SnO2 (and Sn has a much lower MP) and it needs basically red heat to proceed at appreciable rate. When I did this years ago, the
SnO2/C mixture behaved like a 'fluidised bed' because the escaping CO/CO2 caused 'bubbles' to form in the reagent mix. Interesting to watch.
[Edited on 13-6-2013 by blogfast25]MrHomeScientist - 13-6-2013 at 05:30
He used copper carbonate though, does that make a difference?
That's actually my video, and there were several things wrong with it that I'd like to revisit. You saw I did get some copper, but it was really just
a dusting. Two things are key that I didn't do:
1) Powdering the reactants. This provides lots of surface area for the reaction to occur, and is generally a good idea for just about any chemical
reaction.
2) Cover the crucible. As watson.fawkes mentioned, production of CO will also occur. This gas can act as an intermediate which, if trapped by a loose
lid, can react again with the reactants to free more elemental copper.
Changing those two things would have improved my yield quite a bit, I think. My charcoal burner setup doesn't get near to the melting point of copper,
though - I've only just barely melted aluminum with it. A good backyard furnace is something I've wanted for a long time.blogfast25 - 13-6-2013 at 12:13
2) Cover the crucible. As watson.fawkes mentioned, production of CO will also occur. This gas can act as an intermediate which, if trapped by a loose
lid, can react again with the reactants to free more elemental copper.
That would barely, if at all, have made a difference. This reaction is largely driven by the volatility of CO/CO2, which leave the reaction product
mix, thus driving the equilibria:
... to the right. At very high temperatures it's mainly the first reaction that takes place.
Your 'failure' was largely due to far too low temperature. Had you sat this directly onto the red hot coals of a barbecue, with good fanning or
bellowing below the pan, it would have proceeded much better. Powdering and using an excess C also are important... Try it again with a smaller
crucible (and reagent mix), straight onto a blazing BBQ (using lump charcoal, not 'briquettes').
[Edited on 13-6-2013 by blogfast25]DraconicAcid - 13-6-2013 at 12:36
You might be better off heating the copper oxide in a stream of natural gas.Eddygp - 13-6-2013 at 12:57
Well, bear in mind the next reactions.
Being β and λ rational numbers, the reactions A and B are as follow:
A= 2 CuO + C ==> 2 Cu + CO2
B= CuO + C ==> Cu + CO
Your final reaction will be β*A + λ*B so it is
(2β+λ) CuO + (β+λ) C ==> (2β+λ) Cu + β CO2 + λ CO
Control the parameters β and λ somehow to control the reaction or just add a lot of carbon in excess, like 4 times the needed amount.
Pulverize both the CuO and C for best results and if you would like even better results, do it in an inert atmosphere.
Actually, you can add a C reaction that would occur if the reaction B happens:
C= CuO + CO ==> Cu + CO2
Being ω any number, apart from the other two previously defined, we get that:
(2β+λ+ω) CuO + (β+λ) C ==> (2β+λ+ω) Cu + (β+ω) CO2 + (λ-ω) CO
Which of course is absolutely crazy to perfectly work with it but correct to a certain extent. The (λ-ω) CO is because you would put ω
CO on the other side and you can just pass it over, depending on which coefficient is larger.
[Edited on 14-6-2013 by Eddygp]blogfast25 - 14-6-2013 at 03:58
One way of determining whether a metallurgical reduction reaction can proceed and at which temperature is by using Ellingham Diagrams. These plot the
Gibbs Free Energy change ΔG in function of temperature, for the reactions involved in the process.
For the CuO/C system, select ‘oxides’ and tick Cu and C on the periodic table, then ‘see Ellingham Diagram’.
The rules are:
1) If the line for C to CO lies below the oxidation line of the element, then the oxide of the element can be reduced by C, yielding CO. For CuO that
is the case from roughly 500 K onwards.
2) If the line for C to CO2 lies below the oxidation line of the element, then the oxide of the element can be reduced by C, yielding CO2. For CuO
that is the case across the board.
3) If the line for CO to CO2 lies below the oxidation line of the element, then the oxide of the element can be reduced by CO, yielding CO2. For CuO
that is only possible at very high temperature.
Of course it also works for other combinations of oxides/reducing agents, as well as compounds other than oxides.
Try CuO + H2 for instance…
[Edited on 14-6-2013 by blogfast25]Eddygp - 14-6-2013 at 12:35
Actually, I think you could just forget my Freudian slip.Eddygp - 14-6-2013 at 12:35
Actually, I think you could just forget my Freudian slip.blogfast25 - 14-6-2013 at 12:36
Freudian slip? You've truly lost me now...Eddygp - 16-6-2013 at 13:42
