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

AI Thread Summary
To determine the volume of Hg(NO3)2 needed for complete precipitation, the reaction between sodium iodide (NaI) and mercury (II) nitrate (Hg(NO3)2) must be analyzed. The balanced equation indicates a 2:1 molar ratio of NaI to Hg(NO3)2. First, calculate the moles of NaI in the 24.0 mL of 0.170 M solution, which equals 0.00408 moles. Using the molar ratio, 0.00204 moles of Hg(NO3)2 are required, and with its concentration of 0.209 M, the volume needed can be calculated as approximately 9.76 mL. Understanding the stoichiometry of the reaction is essential for solving the problem.
Mackydoodle
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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 :)
 
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Yes there are 2 Moles of NaI in the equation
I believe the mol ratio is 2/1
 
Mackydoodle said:
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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