[gotalbin]

You are given a 1.50 g mixture of sodium nitrate and sodium chloride. You dissolve this mixture into 100 ml of water and then add an excess of .500 Molar silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white sand has a mass of .641 g.

a. List the species you would see if you had a magnified view of this solution
b. White the balanced net ionic equation for the reaction that produces the solid.
c. Calculate the percent sodium chloride in the original unknown mixture.

For question A: I believe the species are Na⁺, Cl^-, and NO₃^-.

I am not sure about question B because I don't know what they mean by net ionic reaction. My guess at what the formula would be is:
NaCl + NaNO₃ ------> NaNO₃ + NaCl

But I don't think that this is the reaction that produces the solid. Actually is it a reaction that produces the solid; isn't the solid just the leftover silver nitrate that didn't react. I'm not sure.

I am pretty much stumpted on Question C. I would appreciate any sort of help on this question. Thanks.

[tinhb]

For question A, i don't know which solution they want to ask so i can't answer
B. The ionic equation : Ag⁺ + Cl^- -> AgCl ( this is the white solid )
C. Mol of sodium chloride = Mol of Cl^- = Mol of white sand = .641 / 143.5
~ .0045 (mol). Then you can finish the problem by yourself

[Padfoot]

a) Yes, but also silver ions - wasn''t the silver nitrate solution in excess?

b) Look at your chemical reaction -what has happened to produce the white sand?  The reaction must have been between silver nitrate and soduim chloride as tinhb says. 

c) tinhb correct - because excess silver nitrate was added.

[gotalbin]

So would the answer to question c be 17.5%.

I did .0045 mol NaCl times its molecular weight to get .263 grams NaCl

Then .263/1.50 times 100 = 17.5%

[Padfoot]

Looks right to me