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25 August, 23:29

How many litres of oxygen at stp can be obtain from 110g of potasium chlorade?

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  1. 26 August, 03:11
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    First, you have to determine the reaction involved between potassium chlorate to oxygen. When potassium chlorate, or KClO₃, is heated, it is decomposed into potassium chloride, KCl, and oxygen gas. The reaction is written below:

    2 KClO₃ ⇒ 2 KCl + 3 O₂

    So, the for every 2 moles of KClO₃ heated, it yields 3 moles of O₂. Let us first convert 110 grams to moles using the molar mass of KClO₃ equal to 122.55 g/mol.

    110 g (1 mol/122.55 g) = 0.8976 mol KClO₃

    Then, we use the stoichiometric ratios in the reaction:

    Amount O₂ produced = 0.8976 mol KClO₃ (3 moles O₂ / 2 moles KClO₃)

    Amount O₂ produced = 1.3464 moles

    Now, we assume the oxygen is ideal gas to be able to use the ideal gas equation: PV = nRT. At STP (standard temperature and pressure), the pressure is equal to 101.325 Pa while the temperature is 273.15 K. Substituting the values:

    (101.325) (V) = (1.3464) (8.314) (273.15)

    V = 30.18 m³

    Since 1 m³ = 1000 L, the amount of volume in liters of oxygen produced is 30,176 L.
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