4. What volume of oxygen (NTP) will...
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4. What volume of oxygen (NTP) will be required to react with no A sample of KC103 on decomposition yielded 448 mL of oxygen gas at NIE. Calculate: (a) Weight of oxygen produced (b) Weight of KClO3 originally taken, and (C) Weight of KCl produced. (K = 39, C1 = 35.5 and 0 = 16) 1 . coriac of recents so as to convert 11 of 1

JEE/Engineering Exams
Chemistry
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( (i) ) ( P vee=n R T ) ( n=frac{p v}{R T} ) ( =0.01864 mathrm{mol}_{2} mathrm{O}_{2} ) weight of 2 produced = ( rightarrow 0+8 ) ( =0.01864 mathrm{motes} times frac{32.0 mathrm{g}}{1 mathrm{mot}} ) ( =0.596 mathrm{g} mathrm{O}_{2} ) ii) ( sigma cdot 01864 ) male ( frac{2 x}{3 operatorname{met} t cdot theta_{2}} ) ( =0.01243 mathrm{mol}_{3} mathrm{RClD}_{3} ) solkuo, ( =frac{0.01243 times 122.559}{text { meter }} ) ( - ) ( =1.52 mathrm{g} times mathrm{a} 0.3 ) ( =0.01243 ) moles ( mathrm{KC} ) ( 0.9729 times 4 )
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