posted by James on .
The traditional method of analyzing the amount of chloride ion present in a sample was to dissolve the sample in water and then slowly add a solution of silver nitrate. Silver chloride is very insoluble in water, and by adding a slight excess of silver nitrate, it is possible to effectively remove all chloride ion from the sample.
Ag+(aq) + Cl-(aq) -> AgCl(s_
Suppose that a 1.054-g sample is known to contain 10.3% chloride ion by mass. What mass of silver nitrate must be used to completely precipitate the chloride ion from the sample? What mass of silver chloride will be obtained?
I'm confused on what to do. What do I do with the silver nitrate?
Ag^+(aq) + Cl^-(aq) ==> AgCl(s)
You are starting with 1.054 x 0.103 = grams Cl^-.
Use stoichiometry to calculate how much Ag^+ is needed, convert that to AgNO3 to know how much AgNO3 is needed. In the preamble the talk is about adding a slight excess, and that is correct, so I would fudge a little more AgNO3 (like rounding 1.8 grams to 2.0 or something like that). In a separate problem, you want to convert moles Cl^- to AgCl then to grams to answer that part of the question.
So how would I put AgNO3 in the equation?
That part kind of confuses me.
AgNO3(aq) + Cl^-(aq) ==> AgCl(s) + NO3^-(aq)
k2cl3 ----> 3o2 + KCl
What mass of silver chloride can be produced from 2.00 L of a 0.183 \it M solution of silver nitrate?