simple gravimetric
 

How many grams of NaCl are required to rpecipitate practically all the Ag ions from 2,5 x 10^2 mL of 0,0113 M AgNO3 Solution?? How about the net ionic equation for the reaction.

How do I it? I know the concepts fine but when the question is modified like this, I can't do it. Thanks for helping me out...I really need to study for my test.
:roll:

NaCl(s) + Ag(+) --> Na(+) + AgCl (s)

now you have the stoichiometry and you can calculate the amount of mol NaCl needed. Then you only need to convert that to gram.

Okay. So I don't get it. It is AgNO3 that has volume and molarity then how is it possible for me calculate it? Where has that NO3 gone?? Sorry for asking too much :(

simple for others
 

You observed in the equation that you have mole/mole stoechiometry. Then your real problem is how many moles of AgNO3 you have in the solution.
The answer is: 2500 mL=2.5Lx0.0113=.. calculate yourself.
The result is exact No of NaCl moles you must use.
The last step is calculating the quantity of NaCl/ grams.
Concerning the "loss of NO3 ions": pls do not complicate the problem, is irrelevant.

You add solid NaCl to a solution containing Ag(+)-ions and NO3(-)-ions. since NaNO3(-) is soluble in water, it isn't necessary to write NO3(-) in the reaction equation, since NaNO3(aq) = Na(+)(aq) + NO3(-)(aq)

So we would get:

NaCl(s) + Ag(+) + NO3(-) --> Na(+) + NO3(-) + AgCl(s)

This is equal to:

NaCl(s) + Ag(+) --> Na(+) + AgCl(s)

Re: simple for others
 

how do I calculate the quantity of NaCl/grams?
Divide the mass with molar mass of NaCl??