InterviewSolution
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InterviewSolution
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A 1 - g sample of KClO_(3) washeated under suchconditions thata part ofit decomposedaccording tothe equation(i) 2KCiO_(3) = 2 KCl + 3O_(2)and theremainingunderwentchange accurdingto the eqation : (ii) 4KCIO_(3)= 3KCIO_(4) + KCIIf theamount of O_(2)evelved was 146.8mL, at NTP, calculate the percentage byweight of KClO_(4) in the residue . |
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Answer» Solution :`KCIO_(3) to KCI + O_(2)` APPLYING POAC for O atoms in the eqn. (i), ,oles of O in `KCIO_(3) = ` moles of O in `O_(2)` `3 xx` moles of `KCIO_(3) = 2XX ` moles of ` O_(2)` `3 xx (wt . if KCIO_(3))/( "mol. wt. of " KCIO_(3)) = 2 xx ("volumeatNTP (mL))/( 22400)` Wt. of `KCIO_(3)= (2 xx 146 . 8 xx 122.5)/( 3 xx 22400)` = 0.5358 g Againapplying POAC for K atoms, moles of K atoms in `KCIO_(3)` = moles of K atoms in KCI `("wt. of " KCIO_(3))/( "mol. wt. of " KCIO_(3)) = ("wt. of KCI")/( "mol. wt. of KCI")` Wt. of KCI ` = (0. 5358)/(122.5) xx 74.5 = 0.3260g"". . . (i)` In the secondreaction : the amount of `KCIO_(3)` left = 1 - 0.5358 = 0.4642 g We have, `KCIO_(3) to KCIO_(4) + KCI` 0.4642 g. Applying POAC for O atoms, moles of O in `KCIO_(3)` = moles ofO in ` KCIO_(4)` `3 xx ` moles of `KCIO_(3) = 4 xx ` molesof `KCIO_(4)` `3 xx ("wt. of " KCIO_(3))/("mol. wt. of "KCIO_(3))= 4 xx ("wt. of " KCIO_(4))/( "mol. wt.of " KCIO_(4))` Wt. of `KCIO_(4) = (3 xx 0.4642 xx 138.5)/( 122.5 xx 4)` = 0.3937 g. . . (ii) WT. of KCIproducedby second reaction = wt. of `KCIO_(3)` - wt. of `KCIO_4)` = 0.4642 - 0.3937 = 0.0705 g. . . (iii) Now since on heating `KCIO_(3),O_(2)` shall ESCAPE out, the substance asresidue are KCIproducedby THEREACTION (i) and (ii) and `KCIO_(4)` Wt. of residue = (i) + (ii) + (iii) = 0.3260 +0.3937 +0.0705 = 0.7902 g `:. %of KCIO_(4)` in the residue `= (0.3937)/(0.7902) xx 100` `= 49.8%` |
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