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8 December, 16:49

Exactly 12.0 grams of potassium chlorate is placed in a reaction vessel; following the decomposition, 1.55g of kClO3 - remain unreacted. Calculate the total volume of oxygen collected at 25 degrees celsius and 0.95atm.

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  1. 8 December, 17:34
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    3.3L

    Explanation:

    Firstly, we write the reaction equation:

    2KClO3 - --> 2KCl + 3O2

    Now we need to know the mass of potassium chlorate reacted = total mass - mass of unreacted chlorate

    = 12.0 - 1.55 = 10.45g

    Now we need to know the number of moles of chlorate reacted. This equals the mass reacted divided by the molar mass.

    The molar mass of potassium chlorate = 39 + 35.5 + 3 (16) = 122.5g/mol

    The number of moles = 10.45/122.5 = 0.0853moles.

    From the reaction equation, we know that 2 moles of chlorate yielded 3 moles of oxygen. The number of moles of oxygen produced from 0.0853moles chlorate is thus = 0.12795moles

    Now, to calculate the total volume of oxygen collected, we use the ideal gas equation.

    Here PV = nRT

    hence, V = nRT/P

    We have the following values: P = 0.95atm

    If 1 atm = 101,325Pa, 0.95atm = 101325 * 0.95 = 96258.75Pa

    T = 25 degrees Celsius = 25 + 273.15 = 298.15K

    n = 0.12795 moles

    R = molar gas constant = 8314.462L. Pa/K. mol

    Putting these values into the equation: V = (0.12795 * 8314.462 * 298.15) / 96,258.75 = 3.3L
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