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Question

A sample containing only CaCO3 and MgCO3 is ignited to CaO and MgO. The mixture of oxides produced weight exactly half as much as the original sample. Calculate the mass percentages of CaCO3 & MgCO3 in this sample.

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Solution

Dear student

Overall reaction is as follows:
CaCO3 + MgCO3 ==> CaO + MgO + 2CO2
Let X = moles of CaCO3
X will also equal the moles of CaO
Let Y = moles of MgCO3
Y will also equal the moles of MgO
Molar mass of CaCO3 = 100 g/mol
Molar mass of CaO = 56 g/mol
Molar mass of MgCO3 = 84.3 g/mol
Molar mass of MgO = 40.3 g/mol
Total mass of reactants = 100X + 84.3Y
Total mass of oxides = 56X + 40.3Y
Mass of oxides is half as much as original sample, so
(2)(56X + 40.3Y) = 100X + 84.3Y
112X + 80.6Y = 100X + 84.3Y
12X = 3.7Y
Now let X = 1 mole and Y must be equal to 3.24 moles
Mass of CaCO3 in sample = 100 g
Mass of MgCO3 in sample = 273 g

Percentage of CaCO3 in sample = (100/373) = 26.8%
Percentage of MgCO3 in sample = (273/373) = 73.2%

Regards


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