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20 December, 03:23

A 1.26-g sample of an alkaline earth metal chloride is treated with excess silver nitrate. All of the chloride is recovered as 3.26 g of silver chloride. Identify the metal.

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  1. 20 December, 03:26
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    The metal is calcium (to form CaCl2)

    Explanation:

    Step 1: Data given

    Mass of sample of an alkaline earth metal chloride = 1.26 grams

    Silvernitrate (AgNO3) = in excess

    All of the chloride is recovered as 3.26 g of silver chloride.

    Step 2: The equation

    XCl2 + 2AgNO3 → X (NO3) 2 + 2 AgCl

    Step 3: Calculate moles AgCl

    Moles AgCl = mass AgCl / molar mass AgCl

    Moles AgCl = 3.26 grams / 143.32 g/mol

    Moles AgCl = 0.0227 moles

    Step 4: Calculate moles XCl2

    For 1 mol XCl2 we'll have 2 moles AgCl

    For 2*0.0227 moles we need 0.0227/2 = 0.01135 moles

    Step 5: Calculate molar mass XCl2

    Molar mass XCl2 = mass / moles XCl2

    Molar mass XCl2 = 1.26 grams / 0.01135 moles

    Molar mass XCl2 = 111 g/mol

    Step 6: Calculate of metal X

    Molar mass X = molar mass XCl2 - molar mass Cl2

    Molar mass X = 111 - 70.9 = 40.1 g/mol

    The metal is calcium (to form CaCl2)
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