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InterviewSolution
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1 g sample of `KClO_(3)` was heated under such conditions that a part of it decomposed according to the equation : `2KClO_(3) rarr 2KCl + 3O_(2)` and the remaining underwent change according to the equation : `4KClO_(3)rarr 3KClO_(4)+KCl` If the amount of `O_(2)` evolved was 146.8 ml at S.T.P., calculate the % by weight of `KClO_(4)` in the reside. |
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Answer» Moles of `O_(2)` evolved `=(146.8)/(22400)=6.55xx10^(-3)` Moles of `KClO_(3)` decomposed in first reaction `=(2)/(3)xx6.55xx10^(-3)=4.36xx10^(-3)` Mass of `KClO_(3)` decomposed in first reaction `= 122.5xx4.36xx10^(-3)=0.534 g` Mole of `KClO_(3)` decomposed in sescond reaction `=(1-0.534)/(122.5)=3.8xx10^(-3)` Moles of `KClO_(4)` formed `=(3)/(4)xx3.8xx10^(-3)=2.85xx10^(-3)` Mass of `KClO_(4)` formed `=2.85xx10^(-3)xx138.5=0.395 g` Total moles of `KCl = 4.36xx 10^(-3)+(1)/(4)xx3.8xx10^(-3)=5.31xx10^(-3)` Mass of `KCl=5.31xx10^(-3)xx74.5=0.395 g` `therefore` % of `KClO_(4)=(0.395)/(0.395+0.395)xx100=50%` . |
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