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12 March, 22:09

Potassium chlorate decomposes to potassium chloride and oxygen. If 20.8 g of potassium chlorate decomposes, how many liters of oxygen will form at STP?

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  1. 13 March, 00:17
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    5.71 L of oxygen will be formed at STP

    Explanation:

    Step 1: Data given

    Mass of potassium chlorate = 20.8 grams

    Molar mass potassium chlorate = 122.55 g/mol

    Step 2: The balanced equation

    2KClO3 → 2KCl + 3O2

    Step 3: Calculate moles KClO3

    Moles KClO3 = mass KClO3 / molar mass KClO3

    Moles KClO3 = 20.8 grams / 122.55 g/mol

    Moles KClO3 = 0.170 moles

    Step 4: Calculate moles O2

    For 2 moles KClO3 we'll have 2 moles KCl and 3 moles O2

    For 0.170 moles KClO3 we'll have 3/2 * 0.170 = 0.255 moles O2

    Step 5: Calculate volume O2

    1 mol = 22.4 L

    0.255 moles = 5.71 L

    5.71 L of oxygen will be formed at STP
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