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A 1.0 g sample of KCIO_3 was heated under such conditions that a part of it decomposed according to the equation: (i) 2KCIO_3to 2KCI+3O_2 and the remaining underwent a change according to the equation (ii) 4KCIO_3to 3KClO_4+ KCI. If the amount of oxygen evolved was 146.8 mL at S.T.R, calculate the percentage by weight of KClO_4 in the residue.

Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/suppose-656311" style="font-weight:bold;" target="_blank" title="Click to know more about SUPPOSE">SUPPOSE</a>, <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> g of `KClO_3` decomposed according to the equation: <br/> `underset(2 xx 122.55 g)(2KClO_(3)) to underset(2 xx 74.55 g)(2KCl) + underset(3 xx 22.4 L "at S.T.P.")(3O_(2))` <br/> Therefore, `1-x`g of `KClO_(3)` decomposed according to the equation, <br/> `underset(4 xx 122.55 g)(4KClO_(3)) to underset(3 xx 138.55 g)(3KClO_(4)) + underset(74.55 g)(KCl)` <br/> From these equations, it is <a href="https://interviewquestions.tuteehub.com/tag/clear-408920" style="font-weight:bold;" target="_blank" title="Click to know more about CLEAR">CLEAR</a> that oxygen is evolved only in the first type of decomposition. <br/> `therefore 2 xx 122.55 g` of `KCIO_3` on decomposition gives `O_2 = 3 xx 22.4` L <br/> `therefore` x gm of `KCIO_3` on decomposition will give: <br/> `O_(2) = (3 xx 22.4)/(2 xx 122.55) xx x` L at S.T.P. <br/> `therefore` The percentage of `KClO_(4)` in the residue <br/> `=0.394/0.790 xx 100 = 49.9`</body></html>


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