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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.
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