[schmitgreg]

I am having trouble combining moles and precipitation reactions. I know the following equation is true:

NaCl(aq) + AgNO₃(aq) -> NaNO₃(aq) + AgCl(s)

What I don't understand is how to combine moles with this.
I don't even know how to come up with an equation as an example, but it would look something like:
If 10 grams of AgCl are precipitated with one liter of NaCl, then how much AgNO₃ was needed to precipitate the Silver Nitrate?

I'm not sure if the amounts make sense, but I just don't understand where to start.

In that case I would say that because it is a 1-to-1 ratio of Silver Nitrate precipitated and AgNO₃ needed, then I would say that you would need 10 grams of AgCl. I don't even know if that makes sense because it is a solution. Plus you would need to know the molarity.

Can someone help me understand this? Thanks!

[ARGOS++]

Dear Schmitgreg;

Why not reading correctly what you have teached yourself?

You tell that from your equation follows that you will have a 1-to-1 ratio; “exactly” for such ratio(s) the term “Moles” was ’created’.
Memorise: By a ratio 1-to-1: 1mol reacts with 1mol; or gives 1mol product (in your case).

As you correctly say to precipitate 1mol AgCl (it’s the product), exactly 1mol AgNO₃ is required.
Conclusion: You have to find out, how much moles are 10.0g AgCl and for that you must first “calculate” what the Molecular Weight (MW) of AgCl is.
And now you know how much moles of AgNO₃ were at begin in your beaker.

The last step should now be easy to be calculated by you (the weight of AgNO₃), as soon as you know the MW of AgNO₃.
I’m right? 
Have you seen; you have nearly all told yourself already, only with a small mistake.

I hope this is of help to you

Good Luck!
                    ARGOS⁺⁺

[constant thinker]

In this case you can pretty much just ignore the information about one liter of NaCl. You can go straight from the 10.0g of AgCl to x g of AgNO₃. You can just use stoich.