12 g of magnesium powder was ignited in a container having 20 g of pure oxygen.After the reaction was over, it was found that 12 g of oxygen was left unreacted.Show that it is according to law of constant proportions.2Mg + O2 ⎯→ 2MgO
Question
12 g of magnesium powder was ignited in a container having 20 g of pure oxygen.After the reaction was over, it was found that 12 g of oxygen was left unreacted.Show that it is according to law of constant proportions.2Mg + O2 ⎯→ 2MgO
Solution
Sure, here is the step by step solution:
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According to the balanced chemical equation, 2 moles of Mg react with 1 mole of O2 to form 2 moles of MgO.
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The molar mass of Mg is 24 g/mol and that of O2 is 32 g/mol. Therefore, 48 g of Mg reacts with 32 g of O2.
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In the given problem, 12 g of Mg is used which is 0.5 moles (12 g / 24 g/mol).
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According to the balanced equation, 0.5 moles of Mg should react with 0.25 moles of O2 (0.5 moles Mg * (1 mole O2 / 2 moles Mg)) which is 8 g of O2 (0.25 moles * 32 g/mol).
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Initially, there was 20 g of O2. After the reaction, 12 g of O2 was left unreacted. Therefore, 8 g of O2 was used in the reaction (20 g - 12 g).
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This is exactly the amount of O2 that should have reacted with 12 g of Mg according to the balanced equation.
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Therefore, the reaction follows the law of constant proportions.
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