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A sample of KCIO_(3)on decomposition yield 448 ml of oxygen gas at NTP calculate (i) weight of oxygen produced, (ii) weight of KCIO_(3)originally taken, and (iii) weight of KCI produced (K = 39, CI = 35.5 and O =16) |
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Answer» Solution : (i) Mole of oxygen = `(448)/( 22400) = 0.02"" ` (RULE3 ) Wt.of oxygen ` = 0 . 0 2 xx 32= 0 . 64 g ""` (Rule 1 ) (ii)`KClO_(3) toKCl+ O_(2)` ApplyingPOAC for O atoms, MOLES of O atomsin `KClO_(3) = ` moles of O atomsin `O_(2)` `3 xx` moles of `KCiO_(3) = 2 xx` moles of `O_(2)` (1 moles of `KCiO_(3)`contains 3 moles of O and 1 mole of `O_(2)` contains2 MOLESOF O) `3 xx (wt. of KClO_(3))/(underset(("Rule 1 "))(mol. wt . of KClO_(3))) = 2 xx ("vol . at NTP (litres )")/(underset(("Rule 3" ))(22.4))` `3 xx (w.t of KClO_(3))/( 122. 5) = 2 xx (0. 448)/( 22 . 4)` W.tof `KCiO_(3) = 1 . 634 g.` (iii)Again applyingPOAC for Katoms, moles of K atomsi `KClO_(3)` = moles of K atoms in KCl or`1 xx ` moles of `KClO_(3) = 1 xx` moles of KCl (1 mole of `KCiO_(3)` contains 1 moles of Kand 1 mole of KCl contains 1 moleof K) ` 1 xx (wt. of KClO_(3))/(mol. wt. of KClO_(3)) = 1 xx (wt . of KCl)/( mol. wt. of KCl)` ` (1 . 634)/( 122.5) = (wt . of KCl)/( 74 . 5 )` Wt . of KCl = 0 . 00037g |
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