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Calculating grams of Precipitate

  1. Oct 3, 2007 #1
    1. The problem statement, all variables and given/known data
    If 70.0 mL of 0.150 M CaCl2 is added to 15.0 mL of 0.100 M AgNO3, what is the mass in grams of AgCl precipitate?

    2. Relevant equations
    Limiting reagants.

    3. The attempt at a solution
    So I tried to find the limiting reagent, but that's where I'm stuck. I'm using examples from my textbook and notes, but everything is in g instead of mL/L. So it's quite confusing. I don't even know how to start.
    I tried:
    (.15M CaCl2)(.07L CaCl2)(1 mol CaCl2/2 mol AgNO3)
    And that's about as far as I got, because I didn't even know if I was doing the right steps.

    Any help is very appreciated.
  2. jcsd
  3. Oct 3, 2007 #2


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    To be more sure, first compare the number of moles of CaCl2 to the number of moles of AgNO3; then, use the ratio information of 1 mole Calcuim Chloride to 2 moles Silver Nitrate to determine which compound is the limiting reactant.

    Moles calcium chloride available = 0.07L * 0.15 M = ( )
    moles silver nitrate available = 0.015L * 0.1 M = ( )

    Which will one will still have a portion unreacted after mixing?
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