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%