one mole of potassium chlorate is thermally decomposed and excess of magnesium is burnt in the obtained gaseous product. The weight of MgO formed will be
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Two reactions occur here: the thermal decomposition of KClO₃ and the subsequent oxidation of Aluminum to form Al₂O₃. Our 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....Here's the math
1.5 mol O₂ x ( 2 mol Al₂O₃ / 3 mol O₂) = 1 mol Al₂O₃
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....Here's the math
1.5 mol O₂ x ( 2 mol Al₂O₃ / 3 mol O₂) = 1 mol Al₂O₃
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