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chemistry 11

Food manufacturers sometimes add calcium acetate to puddings and sweet sauces as a thickening agent. What volume of 0.500 mol/L calcium acetate, Ca(CH3COO)2 (aq), contains 0.300 mol of acetate ions?

So I've been over this a hundred times and I still can't get past the part where i find the moles of calcium acetate.

I can't do

Moles Ca(CH3COO)2= MASS x (1 mol/ 158.11)
158.11 is the molar mass of calcium acetate, by the way

I can't find the moles for calcium acetate, so then i can't do the net ionic equation and do the mole ratio... I'm just really confused about this question

I'm really sorry if I don't make sense, but I'd really appreciate your help

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  1. Follow up if need be.

    You want 0.3 mol of Ac^-. That will be 1/2 that or 0.15 mol of Ca(Ac)2 since 1 mol Ca(Ac)2 contains 2 Ac^- ions.
    Then M = mols/L
    M = 0.5M
    mols = 0.15
    L = ?

    Let me try to do this the way you started. Start with what you have and convert stepwise to end up with what you want.
    0.3 mols Ac^- x (1 mol Ca(Ac)2/2 mol Ac^-) x (1 mol Ca(Ac)2/0.5 mol Ca(Ac)2/L)) = 0.300 L or 300 mL of the 0.5M solution of Ca(Ac)2.
    Note mols Ac^- cancel with mols Ac^- to leave mols Ca(Ac)2 then mols Ca(Ac)2 cancel with mols Ca(Ac)2 to leave L. I prefer to do it in two steps. I think it's easier to follow that way.
    mols Ca(Ac)2 = 1/2 mols Ac^- = 0.30/2 = 0.15.

    Then L = mols M = 0.15/0.5 = 0.300 L or 300 mL.

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