When 5.4 g of NaCl and 3.6 g of AgNO3 are dissolved in water and mixed, what ions remain in the solution and how many moles of each will be there?
NaCl(aq) + AgNO3(aq) ==> AgCl(s) + NaNO3(aq)
mols NaCl = g/molar mass = approx 0.09 but you need to work each step and obtain better accuracy for the answers. All of the following numbers I've made are estimates.
mols AgNO3 = g/molar mass = approx 0.02
The net ionic equation is
Ag^+ + Cl^- ==> AgCl.
Therefore, you can see that the Na^+ and NO3^- are unchanged so beginning mols of those ions and end mols will be the same.
For Ag^+ and Cl^- there is a ppt so those will change.
.........Ag^+ + Cl^- ==> AgCl(s)
I......0.02....0.09......0.00
C.....-0.02...-0.02......0.02
E........0.....0.07.......0.02
So you will have 0.07 mols Cl^- left and "zero" left for Ag^+. The zero probably will be counted as correct for this problem BUT AgCl does have a slight solubility in water and will have a few mols. YOu could calculate how many if the volume were listed.
Summary: 0.09 mols Na^+
0.02 mols NO3^-
0.07 mols Cl^-
almost zero for Ag^+
To determine the ions that remain in the solution and their respective moles, we first need to understand that both NaCl and AgNO3 dissociate into ions when they are dissolved in water.
NaCl dissociates into Na+ and Cl- ions:
NaCl(s) → Na+(aq) + Cl-(aq)
AgNO3 dissociates into Ag+ and NO3- ions:
AgNO3(s) → Ag+(aq) + NO3-(aq)
Let's start by calculating the number of moles of NaCl and AgNO3 using their respective molar masses. The molar mass of NaCl is 58.5 g/mol, and the molar mass of AgNO3 is 169.9 g/mol.
Moles of NaCl = Mass / Molar Mass
Moles of NaCl = 5.4 g / 58.5 g/mol
Moles of NaCl ≈ 0.0923 mol
Moles of AgNO3 = Mass / Molar Mass
Moles of AgNO3 = 3.6 g / 169.9 g/mol
Moles of AgNO3 ≈ 0.0212 mol
Now, let's determine the ions that remain in the solution and their respective moles.
From the dissociation reactions mentioned earlier:
1 NaCl molecule yields 1 Na+ ion and 1 Cl- ion
1 AgNO3 molecule yields 1 Ag+ ion and 1 NO3- ion
Considering the stoichiometry of the compounds, we can conclude that the total number of moles of Na+ ions will be equal to the total number of moles of Cl- ions, and also, the total number of moles of Ag+ ions will be equal to the total number of moles of NO3- ions.
Therefore, in the solution, the ions that remain are Na+, Cl-, and Ag+. Each ion will have the same number of moles as their respective compounds.
Moles of Na+ ions ≈ 0.0923 mol
Moles of Cl- ions ≈ 0.0923 mol
Moles of Ag+ ions ≈ 0.0212 mol
Moles of NO3- ions ≈ 0.0212 mol
So, in the solution, you will have approximately 0.0923 moles of Na+ ions, 0.0923 moles of Cl- ions, 0.0212 moles of Ag+ ions, and 0.0212 moles of NO3- ions.