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anonymous

  • one year ago

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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  1. Shikamaru11
    • one year ago
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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

  2. anonymous
    • one year ago
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    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?

  3. anonymous
    • one year ago
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    @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

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