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

• Mackydoodle
In summary, we are given a reaction between 24.0 mL of .170M sodium iodide and .209M mercury (II) nitrate solution, with the equation Hg(NO3)2+2NaI ----> HgI2+ 2NaNO3. To determine the volume of Hg(NO3)2 needed for complete precipitation, we first need to find the number of moles of NaI in the solution, which can be determined using the ratio of 2:1 in the reaction equation.
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 :)

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 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.

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.

## 1. How do I calculate the volume of Hg(NO3)2 needed for precipitation?

The volume of Hg(NO3)2 needed for precipitation can be calculated using the equation V = m/M, where V is the volume in liters, m is the mass of Hg(NO3)2 in grams, and M is the molar mass of Hg(NO3)2.

## 2. Can the volume of Hg(NO3)2 be measured in any unit?

No, the volume of Hg(NO3)2 should be measured in liters as it is a measure of the amount of space that the substance occupies.

## 3. What is the purpose of using Hg(NO3)2 for precipitation?

Hg(NO3)2 is commonly used in precipitation reactions because it is a strong oxidizing agent and can help to remove impurities from a solution.

## 4. Is there a specific concentration of Hg(NO3)2 that is needed for precipitation?

The concentration of Hg(NO3)2 needed for precipitation will depend on the specific reaction and the desired outcome. It is important to consult a chemical reference or perform a trial experiment to determine the appropriate concentration.

## 5. Are there any safety precautions to consider when working with Hg(NO3)2 for precipitation?

Yes, Hg(NO3)2 is a toxic substance and should be handled with extreme caution. Always wear appropriate personal protective equipment, such as gloves and goggles, when working with this chemical.

• Biology and Chemistry Homework Help
Replies
6
Views
2K
• Biology and Chemistry Homework Help
Replies
2
Views
1K
• Biology and Chemistry Homework Help
Replies
1
Views
3K
• Biology and Chemistry Homework Help
Replies
13
Views
3K
• Biology and Chemistry Homework Help
Replies
4
Views
15K
• Biology and Chemistry Homework Help
Replies
7
Views
2K
• Biology and Chemistry Homework Help
Replies
1
Views
4K
• Biology and Chemistry Homework Help
Replies
2
Views
1K
• Biology and Chemistry Homework Help
Replies
4
Views
2K
• Biology and Chemistry Homework Help
Replies
4
Views
2K