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 ?

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^-)

To answer Part A of the question, you correctly identified the precipitate formed as silver chloride (AgCl).

Now let's move on to Part B, calculating the mass of precipitate formed. To do this, we need to find the limiting reactant.

Step 1: Determine the number of moles of each reactant.
- Moles of silver nitrate (AgNO3): Volume (24.0 mL) x Concentration (2.4 M) = 57.6 mmol = 0.0576 moles
- Moles of sodium chloride (NaCl): Volume (32.0 mL) x Concentration (2.0 M) = 64.0 mmol = 0.0640 moles

Step 2: Use the stoichiometry of the balanced equation to determine the number of moles of the precipitate formed.
From the balanced equation: 1 mole of AgNO3 reacts with 1 mole of NaCl to form 1 mole of AgCl.
Since the stoichiometric ratio is 1:1, the number of moles of AgCl formed will be either 0.0576 moles (from AgNO3) or 0.0640 moles (from NaCl).

Step 3: The reactant that produces the smallest number of moles is the limiting reactant.
In this case, since 0.0576 moles < 0.0640 moles, AgNO3 is the limiting reactant.

Step 4: Use the number of moles of the limiting reactant to calculate the mass of the precipitate formed.
The molar mass of AgCl is 143.32 g/mol.
Mass of AgCl = Moles of AgCl x Molar mass of AgCl
Mass of AgCl = 0.0576 moles x 143.32 g/mol = 8.254 grams

So the mass of precipitate formed is 8.254 grams of silver chloride (AgCl).

Finally, moving on to Part C, calculating the concentration of ions remaining in solution after the precipitate forms.
Since AgCl is insoluble in water, it will form a solid precipitate and be removed from the solution. This means that silver ions (Ag+) and chloride ions (Cl-) will still be in the solution.

1 mole of AgCl produces 1 mole of Ag+ ions and 1 mole of Cl- ions. Therefore, the concentration of Ag+ and Cl- ions remaining in the solution will be the same as the initial concentration of the respective reactant.

The concentration of Ag+ ions remaining in solution is 2.4 M, and the concentration of Cl- ions remaining in solution is 2.0 M.

So, after the formation of AgCl precipitate, the concentrations of Ag+ and Cl- ions remaining in solution are both 2.4 M and 2.0 M, respectively.