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.
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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
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?
@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