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^+
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?
1 answer
1 answer