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A mixture of CaCO3 and MgCO3 weighing 7.85 g was reacted with excess HCL. The reactions are CaCO3+2HCL=CaCl2+H2O+CO2 and MgCO3+2HCl=MgCl2+H2O+CO2. The sample reacted completely to produce 1.94 L CO2 at 25 degrees C and 785 torr. Calculate the percentage of CaCO3 and MgCO3 in the original sample.

  • chemistry -

    Two equations in two unknowns; solve simultaneously.
    Let X = mass CaCO3
    Let Y = mass MgCO3
    ======================
    X + Y = 7.85 grams.
    X(moles mass CO2/molar mass CaCO3) + Y(molar mass CO2/molar mass MgCO3) = grams CO2. [Note: YOu will need to use PV = nRT to calculate moles CO2, then convert that to grams CO2.]
    Solve for X and Y then apply
    %CaCO3 = (mass CaCO3/mass sample)*100 = ??
    %MgCO3 = (mass MgCO3/mass sample)*100 = ??
    Post your work if you get stuck.

  • chemistry -

    Thank you. But could you explain how you got that equation for solving for x and y? I need to prove my answer.

  • chemistry -

    .819 mol CO2= 30604 g CO2
    X= 3.604 g CO2/100.09 g/molCaCO3= .036 mol CaCO3
    Y=3.604 g CO2/84.32 g/mol MgCO3= .043 mol MgCO3

    I'm stuck. I don't think I did it right.

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