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

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Discussion Overview

The discussion revolves around a homework problem involving the precipitation reaction between sodium iodide and mercury (II) nitrate. Participants are trying to determine the volume of mercury (II) nitrate solution required for complete precipitation based on a given concentration and volume of sodium iodide.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant expresses uncertainty about how to start solving the problem and requests help.
  • Another participant asks for the number of moles of sodium iodide in the solution and questions the understanding of the reaction equation.
  • A participant confirms that there are 2 moles of sodium iodide in the balanced equation and suggests a molar ratio of 2:1 for sodium iodide to mercury (II) nitrate.
  • Another participant reiterates the presence of 2 moles of sodium iodide in the equation while clarifying that they are inquiring about the moles in the solution, not just the equation, and emphasizes the importance of understanding the ratio of reactants.

Areas of Agreement / Disagreement

There is no consensus yet, as participants are still exploring the problem and clarifying their understanding of the reaction and the quantities involved.

Contextual Notes

Participants have not yet calculated the actual moles of sodium iodide present in the solution, which is necessary for further progress in 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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