Calcium Chloride reacts with sodium carbonate (aq) to form solid calcium carbonate and aqueous sodium chloride. Determine the volume of a 2.00M Calcium chloride solution that would be needed to exactly react with 0.0750L of 1.00M sodium carbonate. (Use BCA table)

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  1. I have no idea what a BCA table is. BCA might be "beginning", "change", A?

    mols mols Na2CO3 = M x L = 0.075
    ...CaCl2 + Na2CO3 ==> CaCO3 + 2NaCl
    Convert mols Na2CO3 to mols CaCl2 by
    0.075 mols Na2CO3 x (1 mol CaCl2/1 mol CaCl2) = 0.075 x 1/1 - 0.075. So now you know 0.075 mols Na2CO3 will require 0.075 mols CaCl2.
    Then M = mols/L. You know mols CaCl2 and M CaCl2; solve for L

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