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What volume of Hg(NO3)2 is required for precipitation?

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Given that 24.0 mL of .170M sodium iodide reacts with .209M mercury (II) nitrate solution according to the below equation (it is balanced).
    What volume of Hg(NO3)2 is required for complete precipitation?

    2. Relevant equations
    Hg(NO3)2+2NaI ----> HgI2+ 2NaNO3


    3. The attempt at a solution
    I have not idea where to start at all, I'm lost. Help is greatly appreciated thanks :)
     
  2. jcsd
  3. Nov 5, 2012 #2

    Borek

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    Staff: Mentor

  4. Nov 5, 2012 #3
    Yes there are 2 Moles of NaI in the equation
    I believe the mol ratio is 2/1
     
  5. Nov 5, 2012 #4

    Borek

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    Staff: Mentor

    I am not asking about moles in the equation, I am asking about moles in the SOLUTION.

    2:1 or 1:2, depends on ratio of what to what. But that's not a bad start.
     
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