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25 June, 05:07

What volume of O2 collected at 22.0 and 728 mmHg would be produce by the decomposition of 8.15 g KClO3?

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  1. 25 June, 05:22
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    There is 2.52 L of O2 collected

    Explanation:

    Step 1: Data given:

    Temperature = 22.0 °C

    Pressure = 728 mmHg = 728 / 760 = 0.958 atm

    Mass of KClO3 = 8.15 grams

    Molar mass of KClO3 = 122.55 g/mol

    Step 2: The balanced equation

    2KClO3 (s) → 2KCl (s) + 3O2 (g)

    Step 3: Calculate moles of KClO3

    Moles KClO3 = mass KClO3 / molar mass KClO3

    Moles KClO3 = 8.15 grams / 122.55 g/mol

    Moles KClO3 = 0.0665 moles

    Step 4: Calculate moles of O2

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

    For 0.0665 moles of KClO3 we have 3/2 * 0.0665 = 0.09975 moles

    Step 5: Calculate vlume of O2

    p*V = n*R*T

    V = (n*R*T) / p

    ⇒ with n = the number of moles O2 = 0.09975 moles

    ⇒ with R = the gas constant = 0.08206 L*atm/K*mol

    ⇒ with T = 22.0 °C = 273 + 22 = 295 Kelvin

    ⇒ with p = 0.958 atm

    V = (0.09975 * 0.08206 * 295) / 0.958

    V = 2.52 L

    There is 2.52 L of O2 collected
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