How is nitrogen gas passing through condensed water in the U-Tube while reducing metal oxides with ammonia gas?

In my school chemistry book, the experiment about the reduction of metal oxides with ammonia has a diagram of the set up as follows:


[![enter image description here](https://i.stack.imgur.com/p1SjD.jpg)](https://i.stack.imgur.com/p1SjD.jpg)


$$\ce{NH₃ + CuO → 3Cu + 3H₂O + N₂}$$


The tube entering the U-tube from combustion tube A is *not dipped* into the condensed water.


So doesn't that mean that nitrogen ([density: $\pu{1.165 kg/m^3}$](https://www.engineeringtoolbox.com/gas-density-d_158.html)), being lighter than water([density: $\pu{1000 kg/m³}$](https://www.engineeringtoolbox.com/gas-density-d_158.html)) will push the water up on reaching a sufficiently high pressure and the water will flow up through Tube B?


I believe nitrogen gas won't pass through as bubbles as the tube is not under the water level. Nitrogen is also almost totally insoluble in water, hence the nitrogen doesn't dissolve in the water either(which can also be seen by nitrogen gas obtained in the gas jar by downward displacement of water) at the right side of the setup.


**So is there a flaw in the diagram? Or am I missing something?**




No, the experiment as depicted is basically correct, see my comment below, however, on what is likely understated.


Remember ammonia is highly soluble in water, so without Tube 'B', the water in Tube 'C' could backflow.


In reality, the volume of Tube 'B' is much larger than depicted to address backflow issues. I would personally replace it with a large heavy jar that sits in the cold water basin (while a light vessel could problematically float, hence the original design of secured tubing).


Don't believe, see what happens upon removing the backflow trap (aka, Tube 'B') when all the CuO is consumed leaving just highly soluble ammonia gas in the system.


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