How many moles of sodium chloride must be added to an aqueous solution that contains 2.0 moles of silver nitrate

in order to precipitate 0.50 moles of silver chloride?
A) 0.25 mol
B) 0.50 mol
C) 1.0 mol
D) 2.0 mol
E) 1.5 mol

The answer is B. I have no idea what to do. I was just messing around and did some dimensional analysis, but I don't think this will help me get the right answer.

I did this:

(2.0mol AgNO3) (1 mol AgCl/4 mol AgNO3), which gives me the .50 mol AgCl. What am I missing?

Thanks a lot!

You have to look up the solubility of silver nitrate in water. Look that up. Now calculate what the beginning concentration, subtract that from the allowed solubility, and you have the concentration to be added, and given volume, you know the amount you can add.

I'm sorry. I'm still not sure I understand. I know that all compounds with Nitrate are soluble, but I don't get what I need to do.

To solve this problem, you need to use the balanced chemical equation between silver nitrate (AgNO3) and sodium chloride (NaCl) to determine the mole ratios between the reactants and products.

The balanced chemical equation for the reaction between silver nitrate and sodium chloride is:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

From the balanced equation, it can be seen that the ratio of AgNO3 to AgCl is 1:1. That means, for every 1 mole of AgNO3, 1 mole of AgCl is formed.

Given that you want to precipitate 0.50 moles of AgCl, you will need the same amount of AgNO3. Therefore, the answer would be 0.50 mol of sodium chloride (NaCl), which is option B.