My little brother had to do this experiment at school, but couldn't finish it because of a lack of time. My chemistry knowledge isn't as well as it used to be, so I couldn't be much of an help.

There is a test tube (Tube A) filled with $\ce{CuO}$ and urea. With a glass straw, this is connected to a second test tube (Tube B), which is filled with lime water (which is if I recall correctly $\ce{Ca(OH)2}$?)

A quick sketch:

quick sketch

I really wanted to recreate this at home but I sadly don't have the tools for it.

The question here is: what happens when tube A is heated? (What happens to the material in tube A? What happens to tube B? What is left behind in A? etc).

My guess would be that $\ce{CO2}$ would flow from tube A to B, but I am not sure even about that.


Heating a hydrocarbon with $\ce{CuO}$ generally evolves $\ce{CO2}$ and $\ce{H2O}$. Urea is no exception for this rule. The $\ce{CO2}$ evolved will turn lime water milky(this is a test for $\ce{CO2}$ gas).

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    $\begingroup$ Urea is not a hydrocarbon (not that it matters much, though). $\endgroup$ – Ivan Neretin Sep 15 '16 at 6:36

Cupric oxide is easily reduced at low temperatures to copper (~$\pu{250^\circ C}$ by $\ce{H2}$ or $\ce{CO}$), and even without a reducing agent, it disproportionates to cuprous oxide and $\ce{O2}$ at ~$\pu{800 ^\circ C}$ (Cotton and Wilkinson: Advanced Inorganic Chemistry). (In aqueous solution, cupric salts are more stable than cuprous.)

Depending on the ratio of $\ce{CuO}$ to urea and temperature, you could get:

  1. $\ce{3 CuO + CO(NH2)2 -> 3 Cu + CO2 + 2 H2O + N2}$, or

  2. $\ce{6 CuO + CO(NH2)2 -> 3 Cu2O + CO2 + 2 H2O + N2}$

The $\ce{CO2}$ passes into the limewater (dilute solution of $\ce{Ca(OH)2}$) and precipitates $\ce{CaCO3}$. The $\ce{N2}$ just bubbles through. The residue in tube (1) could be finely divided copper (black perhaps with a reddish tinge), or yellow (or red) $\ce{Cu2O}$.

Reaction mixture (2) might be more complex. If reaction mixture (2) is heated slowly, the low temperature reaction (forming $\ce{Cu}$) would occur first, and might go completely, leaving a mixture of $\ce{Cu}$ and $\ce{CuO}$ instead of $\ce{Cu2O}$. It would be interesting to control the temperature and follow the progress of the reaction by the amount of gas produced and see what product remains in test tube (1).

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