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A 10.53 g sample containing a mixture of KCl and KClO3 is heated and 1.53 g of oxygen is liberated. Determine the % by weight of the KClO3 in the original mixture.Group of answer choices

Question

A 10.53 g sample containing a mixture of KCl and KClO3 is heated and 1.53 g of oxygen is liberated. Determine the % by weight of the KClO3 in the original mixture.Group of answer choices

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Solution

Sure, here's how you can solve this problem:

  1. First, you need to know the molar mass of KClO3 and O2. The molar mass of KClO3 is approximately 122.55 g/mol and the molar mass of O2 is approximately 32.00 g/mol.

  2. Next, calculate the moles of O2 that were liberated. This can be done using the formula: moles = mass/molar mass. So, moles of O2 = 1.53 g / 32.00 g/mol = 0.0478 mol.

  3. Since the reaction is 2KClO3 -> 2KCl + 3O2, 2 moles of KClO3 produce 3 moles of O2. Therefore, the moles of KClO3 in the original mixture = (2/3) * moles of O2 = (2/3) * 0.0478 mol = 0.0319 mol.

  4. Convert the moles of KClO3 to grams using the molar mass: mass = moles * molar mass = 0.0319 mol * 122.55 g/mol = 3.91 g.

  5. Finally, calculate the percentage by weight of KClO3 in the original mixture: % KClO3 = (mass of KClO3 / total mass) * 100% = (3.91 g / 10.53 g) * 100% = 37.13%.

So, the percentage by weight of KClO3 in the original mixture is approximately 37.13%.

This problem has been solved

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