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QUESTION

# Why does (or "do"?) Barium Oxide + Sulfuric Acid yield Ozone?

Here's what I got.

Ok, now this is a very interesting question.

First thing first, the reaction involves barium peroxide, "BaO"_2, not barium oxide, which is "BaO".

Now, as far as I know, this reaction is used to produce hydrogen peroxide, "H"_2"O"_2.

If you plan on using dilute sulfuric acid, you should use barium peroxide octahydrate, "BaO"_2 * 8"H"_2"O". The reason behind this is that barium sulfate, "BaSO"_4, which is an insoluble ionic compound, forms on the surface of the peroxide, essentially halting the reaction.

Also, the reaction is performed with an ice-cold acid solution because the low temperature slows down the of hydrogen peroxide to water and oxygen gas.

2"H"_2"O"_text(2(aq]) -> 2"H"_2"O"_text((aq]) + "O"_text(2(g]) uarr

The balanced chemical equation for when this reaction (I'll use anhydrous barium peroxide for the sake of simplicity) is performed at low temperature looks like this

"BaO"_text(2(s]) + "H"_2"SO"_text(4(aq]) -> "BaSO"_text(4(s]) darr + "H"_2"O"_text(2(aq])

At room temperature and catalyzed by potassium iodide, "KI", this reaction should proceed like this

2"BaO"_text(2(s]) + 2"H"_2"SO"_text(4(aq]) stackrel(color(red)("KI")color(white)(aa))(->) 2"BaSO"_text(4(s]) darr + 2"H"_2"O"_text((aq]) + "O"_text(2(g]) uarr

So my guess is that this reaction will produce ozone as a side product, maybe depending on a combination of catalyst, reaction temperature, and concentration of the acid.

I was able to find a YouTube video on this supposed reaction - the narration and the subtitles are in Russian, so that will not be very helpful to students who don't speak it.