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

[Mackydoodle]

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 :)

 

[Borek]

How many moles of NaI in the solution?

Do you know how to read the reaction equation?

 

[Mackydoodle]

Yes there are 2 Moles of NaI in the equation
I believe the mol ratio is 2/1

 

[Borek]

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.

 

[DeeNos]

I would first start by understanding the question and the given information. It seems that we are given a reaction between sodium iodide and mercury (II) nitrate, and we need to determine the volume of mercury (II) nitrate needed for complete precipitation.

To solve this problem, we can use the concept of stoichiometry, which is the quantitative relationship between reactants and products in a chemical reaction. In this case, we can use the balanced equation to determine the mole ratio between mercury (II) nitrate and sodium iodide.

From the balanced equation, we can see that for every 1 mole of mercury (II) nitrate, we need 2 moles of sodium iodide. This means that the mole ratio between Hg(NO3)2 and NaI is 1:2.

Next, we can use the given concentrations of the reactants (0.170M NaI and 0.209M Hg(NO3)2) to calculate the number of moles of each.

For NaI: 0.170 moles/L x 0.0240 L = 0.00408 moles

For Hg(NO3)2: 0.209 moles/L x V = 0.00408 moles (since the mole ratio is 1:2, the number of moles of Hg(NO3)2 must be the same as NaI)

Solving for V, we get V = 0.00408 moles/ 0.209 moles/L = 0.0196 L or 19.6 mL.

Thus, we would need 19.6 mL of Hg(NO3)2 for complete precipitation.

I hope this helps and clarifies the problem for you. If you have any other questions, please feel free to ask.