When $\mathrm{KI}$ is reacted with $\mathrm{O}_3$ under aqueous condition the product formed is
When $\mathrm{KI}$ is reacted with $\mathrm{O}_3$ under aqueous condition the product formed is
- $\mathrm{I}_2 \mathrm{O}_4$
- $\mathrm{I}_2 \mathrm{O}_5$
- $\mathrm{I}_4 \mathrm{O}_9$
- $\mathrm{I}_2$
Solution
$\mathrm{O}_3(\mathrm{~g})+2 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{I}_2(\mathrm{~s})+\mathrm{O}_2(\mathrm{~g})$
Asked in: AP EAMCET 2022 (08 Jul Shift 1)
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