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16 March, 12:39

You could add solid KCl to the solution to precipitate out AgCl (s). What mass of KCl is needed to precipitate the silver ions from 12.0 mL of 0.160 M AgNO3 solution? Express your answer with the appropriate units.

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  1. 16 March, 13:14
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    0.143 g of KCl.

    Explanation:

    Equation of the reaction:

    AgNO3 (aq) + KCl (aq) - -> AgCl (s) + KNO3 (aq)

    Molar concentration = mass/volume

    = 0.16 * 0.012

    = 0.00192 mol AgNO3.

    By stoichiometry, 1 mole of AgNO3 reacts with 1 mole of KCl to form a precipitate.

    Number of moles of KCl = 0.00192 mol.

    Molar mass of KCl = 39 + 35.5

    = 74.5 g/mol

    Mass = molar mass * number of moles

    = 74.5 * 0.00192

    = 0.143 g of KCl.
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