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3 November, 22:05

For the reaction 2kclo3 (s) →2kcl (s) + 3o2 (g) calculate how many grams of oxygen form when each quantity of reactant completely reacts.

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  1. 3 November, 22:34
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    First, we need to get the molar mass of:

    KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol

    KCl = 39.1 + 35.5 = 74.6 g/mol

    O2 = 16*2 = 32 g/mol

    From the given equation we can see that:

    every 2 moles of KClO3 gives 3 moles of O2

    when mass = moles * molar mass

    ∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g

    and the mass of O2 then = 3 mol * 32g/mol = 96 g

    so, 245.2 g of KClO3 gives 96 g of O2

    A) 2.72 g of KClO3:

    when 245.2 KClO3 gives → 96 g O2

    2.72 g KClO3 gives → X

    X = 2.72 g KClO3 * 96 g O2/245.2 KClO3

    = 1.06 g of O2

    B) 0.361 g KClO3:

    when 245.2 g KClO3 gives → 96 g O2

    0.361 g KClO3 gives → X

    ∴ X = 0.361g KClO3 * 96 g / 245.2 g

    = 0.141 g of O2

    C) 83.6 Kg KClO3:

    when 245.2 g KClO3 gives → 96 g O2

    83.6 Kg KClO3 gives → X

    ∴X = 83.6 Kg * 96 g O2 / 245.2 g KClO3

    = 32.7 Kg of O2

    D) 22.4 mg of KClO3:

    when 245.2 g KClO3 gives → 96 g O2

    22.4 mg KClO3 gives → X

    ∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3

    = 8.8 mg of O2
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