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27 March, 16:13

A 5.93 g mixture contains both lithium fluoride, LiF, and potassium fluoride, KF. If the mixture contains 3.94 g fluorine, what is the mass of the KF in the mixture?

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  1. 27 March, 19:27
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    The mass of KF in the mixture is 1.067 grams

    Explanation:

    Step 1: Data given

    Total weight of mixture (LiF+KF) = 5.93 grams

    Mixture contains 3.94 grams of fluorine

    Molar mass K = 39.10 g/mol

    Molar mass F = 19.00 g/mol

    Molar mass KF = 58.10 g/mol

    Molar mass of lithium = 6.94 g/mol

    Step 2: mass of KF and LiF

    Mass KF = X grams

    Mass LiF = 5.97 - X grams

    Step 3: Calculate moles

    Moles KF = x grams / 58.10 g/mol

    Moles LiF = (5.97-X grams) / 25.94 g/mol

    1 mol of KF contains 1 mol of F atoms.

    1 mol of LiF contains 1 mol of F atoms.

    moles of F in KF = moles of KF = x/58.10 g/mol

    moles of F in LiF = moles of LiF = (5.97-x) / 25.94g/mol

    Total moles of Fluorine = (x/58.10) + ((5.97-x) / 25.94)

    Step 4: Calculate mass

    The total weight of Fluorine in sample

    3.94 grams = ((x/58.10) + ((5.97-x) / 25.94)) * 19g/mol

    x = 1.067 grams

    The mass of KF in the mixture is 1.067 grams
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