Why didn't you tell us what you used as the limiting reagent? I could have found the error.
AgNO3 + NaCl ==> AgCl + NaNO3
mols AgNO3 = M x L = 0.0576
mols NaCl = M x L = 0.064
So AgNO3 will form 0.0576 mols AgCl.
NaCl will form 0.064 molg AgCl.
The correct answer in limiting reagent problems is ALWAYS the smaller number so 0.0576 mols AgCl will be formed. Convert to grams by mols x molar mass = 0.p0576 x 143.34 = about 8 g AgCl but that's an estimate.
The net ionic equation is
.........Ag^+ + Cl^- ==> AgCl
I.....0.0576...0.064.....0
C....-0.0576..-0.0576...+0.0576
E.......0......0.0064....0.0576
From the above follow this closely. The Na^+ and NO3^- never changed; therefore, those concentrations never changed. You had 0.0576 mols AgNO3 so the (NO3^-) is 0.0576 mols/(0.024L+0.032L) = ? and the (Na^+) is (0.064/total L) = ?
The (Ag^+) and (Cl^-) are done this way. (Ag^+) = solubility of AgCl in a saturated of AgCl solution that is als0 0.0064 mols/0.056L = 0.114 M in Cl^-
........AgCl(s).==> Ag^+ + Cl^-
I.......solid.......0.....0.114
C.......solid.......x......x
E.......solid.......x.....x+0.114
Then Ksp = (Ag^+)(Cl^-) and look up Ksp.
Ksp = (x)(x+0.114) = and solve for x = (Ag^+) and x+0.114 = (Cl^-)
24.0 mL of 2.4 M silver nitrate is mixed with 32.0 mL of 2.0 M sodium chloride.
( I wrote the balanced equation as AgNo3+NaCl -> NaNo3 + AgCl )
A) Identify the precipitate formed by name and formula.
- I wrote AgCl (silver chloride)
B) Calculate the mass of precipitate formed.
- I used the limiting reactant and got 9.4g AgCl (I'm not sure if that it correct or not)
C) Calculate the concentration of all ions remaining in solution after the precipitate forms.
- This is where i get completely lost, i have no clue what to do ?? Help ?
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