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

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SUMMARY

The discussion centers on calculating the volume of mercury(II) nitrate (Hg(NO3)2) required for complete precipitation when reacting with sodium iodide (NaI). Given 24.0 mL of 0.170 M NaI, the stoichiometry of the balanced reaction indicates a 2:1 molar ratio of NaI to Hg(NO3)2. The participants clarify that understanding the moles of NaI in the solution is essential for determining the required volume of Hg(NO3)2.

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  • Understanding of stoichiometry and molar ratios
  • Knowledge of molarity calculations
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  • Calculate the moles of NaI in the solution using the formula: moles = molarity × volume
  • Learn how to apply stoichiometric ratios to determine reactant volumes
  • Study the concept of precipitation reactions in inorganic chemistry
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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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