What mass of silver chloride can be prepared by the reaction of 160.0 mL of 0.25 M silver nitrate with 110.0 mL of 0.20 M calcium chloride?
Calculate the concentrations of each ion remaining in solution after precipitation is complete. (Enter a 0 if none of the ion remains.)
Cl ‾
NO3‾
Ca2+
These really are difficult problems because they are Ksp, common ion effect, and limiting reagent rolled into one. I didn't use the Ksp data except to estimate the Ag^+ and the small amount of additional Cl^-
mols AgNO3 = M x L = about 0.04.
mols CaCl2 = M x L = about 0.022 but you need to do these and all other calculations to do them more accurately since I've estimated here and there.
2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO)3)2
Which is the limiting reagent?
mols AgNO3 x (2 mol AgCl/2 mols AgNO3) = 0.04 x 2/2 = about 0.04 mols AgCl.
mols CaCl2 x (2 mols AgCl/1 mol CaCl2) = about 0.044
Both 0.04 and 0.044 can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value. Therefore, AgNO3 is the limiting reagent and there will be some CaCl2 left un-reacted.
How much CaCl2 is left unreacted?
0.04 mols AgNO3 x (1 mol CaCl2/2 mol AgNO3) = 0.04 x 1/2 = about 0.02 mols USED which leaves 0.022-0.02 = 0.002.
Let's make a table but I will use millimoles instead of mols..
...2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO3)2
I...40.......22.........0.......0
C..-40......-20........40.......20
E...0.........2........40.......20
volume is 160 mL + 110 = 270 mL.
Note the NO3^- didn't enter into the reaction; therefore, it was 40 mmols to start and finish. Final M = 40mmol/270 mL = ?
Ca didn't enter into the reaction either; therefore, it started at 22 and ended at 22/270 = ?
Cl is a Ksp problem. You must recognize that you have pptd AgCl (40 mmols of it) BUT you have an excess of 4 mmols Cl left unused from the CaCl2 (that's 2 mmols CaCl2 x 2 Cl/mol CaCl2 = 4 mmol Cl).
Cl = 4mmol/270 mL = ?
There will be a small additional amount of Cl from the dissolved AgCl but that amounts to only about 2.4E-8M.
To find the mass of silver chloride that can be prepared, we need to determine the limiting reactant in the reaction between silver nitrate and calcium chloride.
Step 1: Write the balanced equation for the reaction:
2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2
According to the equation, 2 moles of silver nitrate (AgNO3) react with 1 mole of calcium chloride (CaCl2) to produce 2 moles of silver chloride (AgCl).
Step 2: Calculate the number of moles of each reactant:
Moles of AgNO3 = volume (in L) x concentration (in mol/L)
Moles of AgNO3 = 0.160 L x 0.25 mol/L = 0.04 mol
Moles of CaCl2 = volume (in L) x concentration (in mol/L)
Moles of CaCl2 = 0.110 L x 0.20 mol/L = 0.022 mol
Step 3: Determine the limiting reactant:
To find the limiting reactant, compare the mole ratio of AgNO3 to CaCl2. The ratio is 2:1, which means that for every 2 moles of AgNO3, you need 1 mole of CaCl2.
Based on the mole calculations, the moles of AgNO3 (0.04 mol) are greater than the moles of CaCl2 (0.022 mol). This indicates that CaCl2 is the limiting reactant.
Step 4: Calculate the mass of AgCl formed:
To calculate the mass of AgCl formed, use the balanced equation: 2 moles of AgCl = 1 mole of CaCl2.
Moles of AgCl = (moles of CaCl2) x (2 moles of AgCl / 1 mole of CaCl2)
Moles of AgCl = 0.022 mol x (2 mol/1 mol) = 0.044 mol
Now, we can calculate the mass of AgCl using the molar mass of AgCl:
Mass of AgCl = moles of AgCl x molar mass of AgCl
Mass of AgCl = 0.044 mol x 143.32 g/mol = 6.3 g (rounded to one decimal place)
Thus, the mass of silver chloride that can be prepared is 6.3 grams.
Now, let's calculate the concentrations of each ion remaining in solution after precipitation is complete.
For Cl-:
Since calcium chloride is the limiting reactant, all of the chloride ions will be used to form silver chloride precipitate. Therefore, the concentration of Cl- remaining in solution will be 0 M.
For NO3-:
Since silver nitrate is not the limiting reactant, there will still be some nitrate ions left in solution. We need to calculate the moles of nitrate ions remaining and then convert it to concentration.
Moles of NO3- remaining = initial moles - moles used for reaction
Moles of NO3- remaining = (initial volume x initial concentration) - (2 moles of AgNO3 used)
Initial moles of NO3- = volume (in L) x concentration (in mol/L)
Initial moles of NO3- = 0.160 L x 0.25 mol/L = 0.04 mol
Moles of NO3- remaining = 0.04 mol - (2 moles of AgNO3 used)
Moles of NO3- remaining = 0.04 mol - (2 x 0.022 mol) = 0.036 mol
Now, we can calculate the concentration of NO3- remaining:
Concentration of NO3- remaining = Moles of NO3- remaining / final volume (in L)
Concentration of NO3- remaining = 0.036 mol / (0.160 L + 0.110 L) = 0.12 M (rounded to two decimal places)
For Ca2+:
Since calcium chloride is the limiting reactant, all the calcium ions will be used to form the precipitate. Therefore, the concentration of Ca2+ remaining in solution will be 0 M.
Finally, the concentrations of each ion remaining in solution after precipitation is complete are:
Cl-: 0 M
NO3-: 0.12 M
Ca2+: 0 M