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18 October, 11:27

How many grams of potassium fluoride can form if 4.00 grams of potassium are reacted with 3.00 grams of fluorine gas according to the reaction: 2K (s) + F2 (g) → 2KF (s)

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  1. 18 October, 13:19
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    We can for 5.93 grams potassium fluoride

    Explanation:

    Step 1: Data given

    Mass of potassium = 4.00 grams

    Mass of fluorine = 3.00 grams

    Molar mass potassium = 39.10 g/mol

    Molar mass fluorine gas = 38.00 g/mol

    Step 2: The balanced equation

    2K (s) + F2 (g) → 2KF (s)

    Step 3: Calculate moles potassium

    Moles potassium = 4.00 grams / 39.10 g/mol

    Moles potassium = 0.102 moles

    Step 4: Calculate moles F2

    Moles F2 = 3.00 grams / 38.00 g/mol

    Moles F2 = 0.0789 moles

    Step 5: Calculate limiting reactant

    Potassium is the limiting reactant. There will react 0.102 moles

    Fluorine gas is in excess. There will react 0.102 / 2 = 0.051 moles

    There will remain 0.0789 - 0.051 = 0.0279 moles

    Step 6: Calculate moles potassium fluoride

    For 2 moles potassium we need 1 mol fluorine to produce 2 moles potassium fluoride

    For 0.102 moles K we need 0.102 moles KF

    Step 7: Calculate mass KF

    Mass KF = moles KF * molar mass KF

    Mass KF = 0.102 moles * 58.10 g/mol

    Mass KF = 5.93 grams
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