When 12.00 moles of potassium chlorate decomposes, how many dm3 of oxygen are produced at 325K and 188 kPa? 2KClO3 →2KCl + 3O2

Question

When 12.00 moles of potassium chlorate decomposes, how many dm3 of oxygen are produced at 325K and 188 kPa?
2KClO3 →2KCl + 3O2
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Delwyn 3 years 2021-08-14T17:44:04+00:00 1 Answers 14 views 0

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  1. Bo
    0
    2021-08-14T17:46:03+00:00
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    Answer:

    258.71 dm³

    Explanation:

    We’ll begin by calculating the number of mole of O₂ produced by the decomposition of 12 moles of KClO₃. This can be obtained as follow:

    The balanced equation for the reaction is given below:

    2KClO₃ —> 2KCl + 3O₂

    From the balanced equation above,

    2 moles of KClO₃ decomposed to produce 3 moles of O₂.

    Therefore, 12 moles of KClO₃ will decompose to produce = (12 × 3)/2 = 18 moles of O₂.

    Finally, we shall determine the volume of the O₂. This can be obtained as follow:

    Temperature (T) = 325 K

    Pressure (P) = 188 KPa

    Number of mole (n) = 18 moles

    Gas constant (R) = 8.314 KPa.dm³/Kmol

    Volume (V) =?

    PV = nRT

    188 × V = 18 × 8.314 × 325

    188 × V = 48636.9

    Divide both side by 188

    V = 48636.9 / 188

    V = 258.71 dm³

    Thus, 258.71 dm³ of oxygen were obtained from the reaction.

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