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$\ce{KO2 + CO2}$ gives $\ce{K2CO3 + O2}$

I think it should be from $\ce{KO2}$ as $\ce{2KO2}$ can split to $\ce{K2O}$ and $\ce{3/2O2}$. This $\ce{K2O}$ being unstable reacts with $\ce{CO2}$ to form $\ce{K2CO3}$.

Only problem is, is the first step possible? Is $\ce{O2}$ released from $\ce{KO2}$ or $\ce{CO2}$?

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    $\begingroup$ Homework is a deprecated tag! $\endgroup$ – Martin - マーチン Dec 20 '17 at 6:55
  • $\begingroup$ KO2 can decompose, but as Nilay Ghosh said, the product is not K2O; also, even that requires heating, which is simply not there. Also, K2O is perfectly stable by itself (though it would surely react with CO2, given a chance). Other than that, you are right. $\endgroup$ – Ivan Neretin Dec 20 '17 at 7:07
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Yes, the oxygen indeed comes from the decomposition of potassium superoxide ($\ce{K2O}$) but the other decomposition product is potassium peroxide ($\ce{K2O2}$) and not potassium monoxide($\ce{K2O}$).(Here)

$$\ce{2KO2 ->[290°C, vacuum] K2O2 + O2~~~~~~~~~~(1)}$$ Now, this potassium peroxide reacts with carbon dioxide to form potassium carbonate and oxygen.(Here)

$$\ce{2K2O2 + 2CO2 → 2K2CO3 + O2~~~~~~~(2)}$$

Adding $\ce{(1)\times2}$ and $\ce{(2)}$, we get full reaction. (Here)

$$\ce{4KO2 + 2CO2 → 2K2CO3 + 3O2}$$


If reaction proceeds in aqueous medium, it gives potassium hydrocarbonate and oxygen.(Here)

$$\ce{4KO2 + 4CO2 + 2H2O → 4KHCO3 + 3O2}$$

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    $\begingroup$ chemiday.com is by no means a reliable source of information. $\endgroup$ – andselisk Dec 20 '17 at 10:00
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    $\begingroup$ "If the reaction proceeds in an aqueous medium"? Doesn't potassium oxide react violently with water? $\endgroup$ – Slarty Dec 20 '17 at 12:54

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