What mass of silver chloride can be prepared by the reaction...

What mass of silver chloride can be prepared by the reaction of 150.0 mL of 0.22 M silver nitrate with 150.0 mL of 0.21 M calcium chloride? Mass = [ ] g Calculate the concentrations of each ion remaining in solution after precipitation is complete. Assume that AgCl is completely insoluble. Concentration of Ag+ = [ ] M Concentration of NO3- = [ ] M Concentration of Ca2+ = [ ] M Concentration of Cl- = [ ] M

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00:01 We want to know what is going to be the mass of silver chloride that is going to be prepared by the reaction of 150 milliliters of 0 .22 molar silver nitrate with the same volume of 0 .21 molar calcium chloride.

00:14 So in this case, the reaction that we're going to be dealing with is going to be silver nitrate, agno3, plus calcium chloride, ccl2.

00:27 Both of them in aqueous solution are going to produce agcl plus calcium nitrate.

00:41 So to balance this reaction, we're going to need to add a 2 over here and a 2 over here just to have our balanced reaction.

00:48 Reaction.

00:48 So the net ionic equation for this is going to be 2 of silver plus plus 2 of cl minus is going to produce 2 of agcl as a solvent.

01:10 So notice that the concentration of no3 minus and the concentration of calcium plus will not change.

01:24 This is going to be the same as the the initial concentrations and in this case the only thing that happened is that the volume doubled so the concentration is going to decrease to one half of the original one so in the case for the nitrate it is going to be 0 .11 instead of 0 .22 because we have 150 and 150 milliliters and 0 .11 for nitrate so 0 .11 molar and for calcium is going to be 0 .105 one half of 0 .21 so with that we're out of the way for both of those results.

02:06 Now, in order to know the concentrations of silver plus and cl minus in solution, we need to determine which one is going to be the limiting reagent.

02:15 So we're going to have the amount of mol of silver plus that we're going to have in solution is going to be equal to the concentration times the volume in liters that we're going to get.

02:28 So, concentration times volume concentration and molarity times volume in liters...

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