How many grams of NaCl are required to precipitate most of the Ag+ ions from 2.50 × 102 mL of 0.0113 M AgNO3 solution? Write the net ionic equation for the reaction.

Question

How many grams of NaCl are required to precipitate most of the Ag+ ions from 2.50 × 102 mL of 0.0113 M AgNO3 solution?  Write the net ionic equation for the reaction.

shikamaru11 · Answer

AgNO3 + NaCl ----> NaNO3 + AgCl(s) 

Ag^1 + NO3^1- + Na^1 + Cl^- ----> Na^1 +NO3^- + AgCl(s) 


now that you are given the amount of mL and the Molarity you can find the amount of moles of AgNO3 

M = mol / Volume (L) 
.0113M = x / .250L 

x = .002825 mol AgNO3 

you now want to set up a molar ratio 

.002825 mol AgNO3( 1 Mol NaCl / 1 mol AgNO3) 

you now have .002825 mol NaCl 

now you can convert .002825 mol NaCl to grams 

.002825 mol NaCl (  NaCl / 1 mol NaCl) 
= .165g NaCl

anonymous · Answer

were did you get the .250L from and also why didn't you separate AgCl on the other side of the equation when you were given the ionic charge?

anonymous · Answer

@shikamaru11 this bottom portion doesn't make seance.  How did you end up with .165g?  What did you do to achieve that number?
.002825 mol NaCl (  NaCl / 1 mol NaCl) = .165g NaCl