what mass of silver chloride can be prepared by the reaction of 100 ml 0,20 M of silver nitrate with 100,0 ml of 0.15 M calcium chloride calculate the concentration of each ion reamaining in solution after the precipitation is complete

This is a limiting reagent (LR) problem wrapped up in a Ksp problem wrapped up in a common ion problem. First calculate how much AgCl we can get and identify the LR.

2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO3)2

millimols AgNO3 = mL x M = 100 x 0.2 = 20
mmols CaCl2 = mL x M = 100 x 0.15 = 15

mmols AgCl formed IF we used the AgNO3 and all of the CaCl2 we needed.
20 x (2 mols AgCl/2 mols AgNO3) = 20

mmols AgCl formed IF we used the CaCl2 and all of the AgNO3 we needed.
15 x (2 mols AgCl/1 mol CaCl2) = 30.
You see we obtained different numbers; in LR problems the smaller value is ALWAYS the correct one and the reagent providing that number is the LR. So AgNO3 is the LR and we form 20 mmols AgCl. You can convert that to g by g = mols x molar mass = ?

How much remains in solution. We didn't touch the NO3^- and we have 2*20 = 40 mmoles in 200 mL so (NO3^-) = mmols/mL = ?
We didn't touch the Ca^2+. That concn is
mmols/mL = 30/200 = ?

...2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO3)2
I...20......15..........0........0
C..-20.....-10.........+20.....+10
E...0.......5..........20(pptd)...10

So we have 5 mmols CaCl2 not used which is 10 mmols Cl^- not used and that is the common ion for the solubility of AgCl.
..........AgCl ==> Ag^+ + Cl^-
I........solid.....0.......0
C........solid.....x.......x
E........solic.....x.......x
Ksp AgCl = (Ag^+)(Cl^-) = Look up Ksp.
Substitute (Ag^+) = x
and (Cl^-) = (x + 10/200) = (x+0.05)
Solve for x to find (Ag^+). The (Cl^-) will be 0.05 + whatever Ag^+ is but that's a small number and probably can be ignored and just call Cl = 0.05
Note: The equation Ksp = (Ag^+)(Cl^-) = Ksp = (x)(x+0.05) is a quadratic if you solve it exactly BUT usually x is small and you can assume x + 0.05 = 0.05. That keeps the quadratic out of the way. I don't know how advanced this course is but the author of the problem MAY NOT have intended to have you go through this last part on the Ksp. Post your work if you get stuck.

To calculate the mass of silver chloride formed, you need to first determine the limiting reactant in the reaction. The limiting reactant is the one that is completely consumed and determines the amount of product formed.

We start by calculating the number of moles of silver nitrate and calcium chloride present:

Moles of silver nitrate = concentration (0.20 M) * volume (100 ml) / 1000 (to convert ml to liters)
= 0.20 * 0.100 / 1000
= 0.02 moles

Moles of calcium chloride = concentration (0.15 M) * volume (100.0 ml) / 1000
= 0.15 * 0.100 / 1000
= 0.015 moles

Next, we need to find the stoichiometric ratio between silver chloride and calcium chloride in the balanced chemical equation. The balanced equation for the reaction between silver nitrate and calcium chloride is:

2AgNO3 + CaCl2 -> 2AgCl + Ca(NO3)2

From the equation, we can see that 2 moles of silver nitrate react with 1 mole of calcium chloride to produce 2 moles of silver chloride.

To determine the limiting reactant, we compare the moles of silver nitrate and calcium chloride in a 2:1 ratio:

Moles of calcium chloride * (2 / 1) = 0.015 * 2
= 0.03 moles

Since the mole ratio is 2:1, we see that 0.03 moles of calcium chloride is greater than the 0.02 moles of silver nitrate.

Therefore, the limiting reactant is silver nitrate, and the amount of silver chloride formed is determined by the moles of silver nitrate.

To calculate the mass of silver chloride, we use the molar mass of AgCl:

Molar mass of AgCl = atomic mass of Ag (107.87 g/mol) + atomic mass of Cl (35.45 g/mol)
= 143.32 g/mol

Mass of silver chloride = moles of silver chloride * molar mass of AgCl
= 0.02 * 143.32
= 2.8664 g (rounded to 2 decimal places)

To calculate the concentration of each ion remaining in solution after the precipitation is complete, we need to determine the total volume of the solution after mixing.

The total volume of the solution is the sum of the initial volumes of silver nitrate and calcium chloride, which is 100 ml + 100.0 ml = 200.0 ml.

First, let's calculate the moles of Ag+ and Cl- initially present:

Moles of Ag+ initially = concentration of Ag+ * volume of Ag+ solution / 1000
= 0.20 * 0.100 / 1000
= 0.02 moles

Moles of Cl- initially = concentration of Cl- * volume of Cl- solution / 1000
= 0.15 * 0.100 / 1000
= 0.015 moles

Since we have a 2:1 stoichiometric ratio between Ag+ and Cl- in AgCl, the moles of Cl- initially present will be consumed completely, resulting in no Cl- ion remaining in solution.

After complete precipitation, the moles of AgCl formed will be equal to the moles of Ag+ initially present. Therefore, the concentration of Ag+ remaining in solution is given by:

Concentration of Ag+ remaining = moles of Ag+ remaining / volume of final solution

The volume of the final solution is the sum of the initial volumes, which is 200.0 ml.

Moles of Ag+ remaining = Moles of Ag+ initially - moles of AgCl formed
= 0.02 - 0.02
= 0 moles

Concentration of Ag+ remaining = 0 / 200.0
= 0 M

Thus, the concentration of Ag+ remaining in solution after the precipitation is complete is 0 M.