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2 August, 13:38

How many grams of oxygen are formed when 58.6 g of kno3 decomposes?

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  1. 2 August, 17:17
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    The formula for potassium nitrate is: 2KNO₃ → 2KNO₂ + O₂

    If moles = mass : molar mass

    then moles of KNO₃ = 58.6 g : [ (39 * 1) + (14 * 1) + (16 * 3) ] g/mol

    = 58.6 g : (101 g) g/mol

    = 0.5802 mol

    Now the mole ratio of KNO₃ : O₂ is 2 : 1

    ∴ if moles of KNO₃ = 0.5802 mol

    then moles of O₂ = 0.5802 mol : 2

    = 0.2901 mol

    Now since mass = moles * molar mass

    ∴ the mass of oxygen produced = 0.2901 mol * (16 * 2) g/mol

    = 9.283 g

    ∴ the grams of oxygen gas produced when 58.6 grams of potassium nitrate is decomposed is ~ 9.28 g
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