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25 April, 01:02

A 0.750-g sample of a compound that might be iron (II) chloride, nickel (II) chloride, or zinc chloride is dissolved in water, and 22.40 mL of 0.515 M AgNO3 is required to completely precipitate all the chloride ion as AgCl. Name the compound, and write its formula.

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  1. 25 April, 01:26
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    Answer: The compound is nickel (II) chloride.

    NiCl2

    Explanation:

    Let's assume that the compound is X.

    Then,

    XCl2 + 2AgNO3 - --> X (NO3) 2 + 2AgCl

    Given,

    Volume of AgNO3 = 22.4ml = 0.0224L

    Concentration of AgNO3 = 0.515M

    no of moles = concentration x volume

    :• n (AgNO3) = 0.515 x 0.0224

    n (AgNO3) = 0.011536 mol

    Therefore, no of moles of AgNO3 is 0.0011536mol

    From the equation above, we can deduce that;

    n (XCl2) = 1/2n (AgNO3)

    :• n (XCl2) = 1/2 * 0.011536

    n (XCl2) = 0.005768

    Therefore number of moles of XCl2 is 0.005768 mol.

    Mass, m, of XCl2 = 0.75g (Given)

    Molar mass = mass / no of moles

    Molar mass of XCl2 = 0.75/0.005768

    :• molar mass of XCl2 = 130.03g/mol

    Molar mass of Cl = 35.453

    Cl2 = 2 x 35.453 = 70.906

    Molar mass of XCl2 = M. mass of X +

    M. mass of Cl2

    :• 130.03 = M. mass of X + 70.906

    Molar mass of X = 130.03 - 70.906

    Molar mass of X = 59.1

    By comparing to literature, we can conclude that X is Nickle, Ni.

    Therefore, the compound is NiCl2, Nickle (II) Chloride.
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