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The thermal decomposition of KClO3 and the subsequent oxidation of Aluminum to form Al2O3.
Two reactions are: 2 KClO3 → 2 KCl + 3 O2 and 4 Al + 3 O2 → 2 Al2O3 The molar ratio for the first reaction is 2:3 for KClO3: O2. Thus, if 1 mole KClO3 is decomposed, 1.5 moles of O2 is produced ===> 1 mol KClO3 x ( 3 mol O2 / 2 mol KClO3) = 1.5 mol O2 The molar ratio for the second reaction is: 3:2 for O2 : Al2O3. ===> 3 moles of oxygen produces 2 moles of aluminum oxide.
Since we have 1.5 moles of oxygen (half of the three needed) we can expect to produce half of the 2 moles or 1 mole aluminum oxide. Hence, 1.5 mol O2 x ( 2 mol Al2O3 / 3 mol O2) = 1 mol Al2O3
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