Silver chloride can be prepared by the reaction of 100mL of 0.20M silver nitrate with 100mL of 0.15M calcium chloride. after the reaction goes to completion, what concentration of which ion remains in solution?
CaCl2 + 2AgNO3 ==> Ca(NO3)2 + 2AgCl
moles CaCl2 = M x L = ??
moles AgNO3 = M x L = ??
Go through the stoichiometry based on the coefficients in the balanced equation to see which reactant has some remaining unreacted. Post your work if you get stuck.
2AgNO3+CaCl2===>2AgCl+Ca(NO3)2 is the balanced equation.
so 0.2 x 0.1 = 0.2mol
= 0.2 (107.9*35.5)= 28.7g
0.15 x 0.1 = 0.015mol
0.015 (110.98)= 1.6647
??
or is it 0.20M x 0.1= 0.02M AgCl
0.15 x 0.01= 0.015M Cl
Neither. You corrected one error from the first post to the second but not the other.
moles AgNO3 = 0.2M x 0.1L = 0.02 moles. Convert to moles AgCl using the coefficients in the balanced equation.
0.02 moles AgNO3 x (2 moles AgCl/2 moles AgNO3) = 0.02 x (2/2) = 0.02 moles AgCl. No need to convert to grams.
moles CaCl2 = 0.15M x 0.1L = 0.015 moles.
Convert to moles AgCl. 0.015 moles CaCl2 x (2 moles AgCl/1 mole CaCl2) = 0.015 x (2/1) = 0.03 moles AgCl.
In limiting reagent problems like this the correct answer is ALWAYS the smaller value and the reagent producing that value is called the limiting reagent. So AgNO3 is the limiting reagent, we obtain 0.02 moles AgCl as the theoretical yield and we have how much CaCl2 that didn't react? I prefer to break this up into ions at this point to keep from getting mixed up.
CaCl2 we have 0.015 mole Ca+2 and 0.03 mole Cl-.
AgNO3 we have 0.02 mole Ag+ and 0.02 mole NO3-.
So all of the Ca+2 is left intact; M = 0.015moles/0.200L = ??M
Cl- = 0.030-0.02 = 0.01 and that divided by 0.2 L.
Ag+ is gone (but you CAN calculate the solubility of AgCl from Ksp).
NO3- = unchanged from original = 0.02M BUT it has been diluted from 100 mL to 200 mL and the final concn is 0.02mole/0.2L = ??
I hope this hasn't confused you. Check my work. It's getting late where I am.
To find out the concentration of the remaining ions in the solution, we need to determine which reaction occurs and the stoichiometry of that reaction. From the reactants given, we can determine that the reaction between silver nitrate (AgNO3) and calcium chloride (CaCl2) will occur, resulting in the formation of silver chloride (AgCl) and calcium nitrate (Ca(NO3)2).
The balanced chemical equation for this reaction is:
AgNO3 + CaCl2 → AgCl + Ca(NO3)2
From the equation, we can see that one mole of silver nitrate reacts with one mole of calcium chloride, producing one mole of silver chloride and one mole of calcium nitrate.
To determine the concentration of the remaining ions in the solution, we need to first calculate the moles of silver nitrate and calcium chloride used in the reaction.
Moles of AgNO3 = volume (L) × concentration (M)
= 0.1 L × 0.20 M
= 0.02 moles
Moles of CaCl2 = volume (L) × concentration (M)
= 0.1 L × 0.15 M
= 0.015 moles
Since the stoichiometry of the balanced equation is 1:1 for silver nitrate and calcium chloride, and assuming the reaction goes to completion, 0.015 moles of each reactant will be consumed.
The reaction produces an equal number of moles of silver chloride and calcium nitrate. Therefore, 0.015 moles of silver chloride are formed.
To find the concentration of remaining ions, we need to calculate the volumes of solutions after the reaction. Since equal volumes of 100 mL (0.1 L) are mixed, the total volume of the solution remains the same, which is 0.2 L.
Now, we can calculate the concentration of the remaining ions in solution:
Concentration of Ag+ ions = moles of Ag+ ions / volume (L) = 0.02 moles / 0.2 L = 0.10 M
Concentration of Ca2+ ions = moles of Ca2+ ions / volume (L) = 0.015 moles / 0.2 L = 0.075 M
Therefore, after the reaction goes to completion, the concentration of silver ions (Ag+) remaining in solution is 0.10 M, and the concentration of calcium ions (Ca2+) remaining in solution is 0.075 M.