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12 December, 10:52

2KCIO3 - - - 2 KCI + 3O2

I want to produce 5.7 mol of O2. How many grams of KCIO3 should i start with?

3.8g KCIO3

698.54g KCIO3

5.7g KCIO3

465.69g KCIO3

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Answers (2)
  1. 12 December, 12:09
    0
    We have to start with 465.69 grams of KClO3 (option 4 is correct)

    Explanation:

    Step 1: Data given

    Moles of O2 = 5.7 moles

    Molar mass KClO3 = 122.55 g/mol

    Step 2: The balanced equation

    2KCIO3 → 2KCI + 3O2

    Step 3: Calculate moles KClO3

    For 2 moles KClO3 we'll have 2 moles KCl and 3 moles O2

    For 5.7 moles O2 we nee 2/3 * 5.7 = 3.8 moles KClO3

    Step 4: Calculate mass of KClO3

    Mass KClO3 = moles KClO3 * molar mass KClO3

    Mass KClO3 = 3.8 moles * 122.55 g/mol

    Mass KClO3 = 465.69 grams

    We have to start with 465.69 grams of KClO3 (option 4 is correct)
  2. 12 December, 14:22
    0
    465.69g KCIO3

    Explanation:

    See the stoichiometry in the reaction:

    We can propose

    3 moles of oxgen are made of 2 moles of chlorate

    Therefore 5.70 moles of O₂ will be made by (5.70. 2) / 3 = 3.8 moles of chlorate.

    We convert the moles to mass: 3.8 mol. 122.55 g / 1 mol = 465.69g KCIO3

    is the mass to use in the begining
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