Calculate the % Na2CO3 and % BaCO3 in a 0.2005g sample which requires 28.0mL of 0.9990 N acid for complete neutralization.

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asked by Summer
  1. You need two equations and solve them simultaneously. The problem doesn't say which acid is used (and it doesn't need to do that) but I will assume HCl and work accordingly. Any acid can be used as long as you keep the N and M straight. The first equation is
    g Na2CO3 + g BaCO3 = total mass in g.
    The second equation is
    mols HCl from Na2CO3 + mols HCl from BaCO3 = total mols HCl.

    BaCO3 + 2HCl ==> BaCl2 + H2O + CO2
    Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2

    Let X = mass Na2CO3
    and Y = mass BaCO3
    Then equation 1 is
    X + Y = 0.2005

    Equation 2 is
    (2X/molar mass Na2CO3) + (2Y/molar mass BaCO3) = mols HCl = 0.028*0.9990

    Solve the two equations for X and Y. Then
    %Na2CO3 = (mass Na2CO3/0.2005)*100 = ?
    %BaCO3 = (mass BaCO3/0.2005)*100 = ?
    Post your work if you get stuck.

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  2. How to know their mass? Is is their molecular mass?

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    posted by Summer

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