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28 January, 12:07

Determine the calcium carbonate equivalent (CCE) of the following compounds: (amount that has the same neutralizing value as 100 g pure CaCO3) (a) MgO (b) Mg (OH) 2 (c) and CaMg (CO3) 2.

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  1. 28 January, 12:50
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    a) 40 g

    b) 58 g

    c) 184 g

    Explanation:

    The calcium carbonate equivalent of any compound is calculated by calculating it molar mass.

    The molar mass of the compound is the calcium carbonate equivalent as it corresponds to 1 mole of the compound and equals to one mole of calcium carbonate then.

    a) MgO

    Atomic mass of Mg = 24

    Atomic mass of O = 16

    Molar mass = CCE = 24+16 = 40 g

    b) Mg (OH) ₂

    Atomic mass of Mg = 24

    Atomic mass of O = 16

    Atomic mass of H = 1

    Molar mass of Mg (OH) ₂ = CCE = 24 + (2X16) + 2 (X1) = 58g

    c) CaMg (CO₃) 2

    Atomic mass of Mg = 24

    Atomic mass of O = 16

    Atomic mass of C = 12

    Atomic mass of Ca = 40

    Molar mass = CCE = 40 + 24 + (2X12) + (6X16) = 184 g.
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