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

  • #1

Homework Statement



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?

Homework Equations


Hg(NO3)2+2NaI ----> HgI2+ 2NaNO3


The Attempt at a Solution


I have not idea where to start at all, I'm lost. Help is greatly appreciated thanks :)
 

Answers and Replies

  • #3
Yes there are 2 Moles of NaI in the equation
I believe the mol ratio is 2/1
 
  • #4
Borek
Mentor
28,441
2,838
Yes there are 2 Moles of NaI in the equation
I am not asking about moles in the equation, I am asking about moles in the SOLUTION.

I believe the mol ratio is 2/1
2:1 or 1:2, depends on ratio of what to what. But that's not a bad start.
 

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