KO_2 + CO_2 + H_2O xrightarrow{text{more } CO | vedclass

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(B) The reaction of potassium superoxide $(KO_2)$ with carbon dioxide $(CO_2)$ and water $(H_2O)$ proceeds as follows:$4KO_2 + 2CO_2 + 2H_2O \rightarr

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$KHCO_3, O_2$

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(B) The reaction of potassium superoxide $(KO_2)$ with carbon dioxide $(CO_2)$ and water $(H_2O)$ proceeds as follows:$4KO_2 + 2CO_2 + 2H_2O \rightarrow 4KHCO_3 + 3O_2$In the presence of excess $CO_2$,the primary product formed is potassium bicarbonate $(KHCO_3)$ and oxygen gas $(O_2)$ is released.Therefore,$[X] = KHCO_3$ and $[Y] = O_2$.

$KO_2 + CO_2 + H_2O \xrightarrow{\text{more } CO_2} [X] + [Y]$
Products $[X]$ and $[Y]$ are respectively

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$KO_2 + CO_2 + H_2O \xrightarrow{\text{more } CO_2} [X] + [Y]$Products $[X]$ and $[Y]$ are respectively