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6 May, 18:51

When heated a sample consisting of only CaCO3 and MgCO3 yields a mixture of CaO and Mgo. If the weight of the combined oxides is equal to 51.00% of the initial sample weight, calculate the per cent of CaCO3 and MgCO3 in the sample. (12 M.)

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  1. 6 May, 22:14
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    38.83 % of CaCO3

    61.17 % of MgCO3

    Explanation:

    where Moles of CaCO3 is equals to x and MgCO3 is y we have that ...

    CaCO3 molar mass = 100.09 g / mol = 100.09 x

    MgCO3 molar mass = 84.31 g / mol = 84.31 y

    decomposition reactions:

    CaCO3 - --> CaO + CO2

    MgCO3 - --> MgO + CO2

    So we have that, Moles of CaO = Moles of CaCO3 = x

    and Moles of MgO = Moles of MgCO3 = y

    CaO molar mass = 56.08 g / mol

    MgO molar mass = 40.30 g / mol

    CaO = 56.08 x

    MgO = 40.30 y

    "If the weight of the combined oxides is equal to 51.00% of the initial sample weight,"

    total mass of MgO and CaO = 51.00 % of Total Mass of MgCO3 and CaCO3

    thus

    56.08 x + 40.30 y = 0.51 (100.09 x + 84.31 y)

    56.08 x + 40.30 y = 51.04 x + 42.99y

    5.04 x = 2.7 y

    y = 1.87 x

    CaCO3 % in the sample

    = 100.09 x * 100 / (100.09 x + 84.31 y)

    = 10009 x / (100.09 x + 84.31 * 1.87 x)

    = 10009 x / (x (100.09 + 157.66))

    = 10009 / 257.75

    = 38.83 %

    MgCO3 % in the sample

    = 100 - 38.83

    = 61.17 %
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