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5 July, 07:07

8. How many liters of oxygen gas will be produced by the decomposition of

100.0 grams of sodium nitrate? (Note: 1 mole of any gas at STP will have a

vol = 22.4 liters.)

2 NaNO3 → 2NaNO2 + O2

+4
Answers (1)
  1. 5 July, 07:12
    0
    13.44 L

    Explanation:

    Given dа ta:

    Volume of oxygen produced = ?

    Mass of sodium nitrate = 100 g

    Solution:

    Chemical equation:

    2NaNO₃ → 2NaNO₂ + O₂

    Number of moles of sodium nitrate:

    Number of moles = mass / molar mass

    Number of moles = 100 g / 85 g/mol

    Number of moles = 1.2 mol

    Now we will compare the moles of NaNO₃ with oxygen

    NaNO₃ : O₂

    2 : 1

    1.2 : 1/2 * 1.2 = 0.6 mol

    Volume:

    one mole = 22.4 L

    0.6 mol * 22.4 = 13.44 L
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