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?
So how would I put AgNO3 in the equation?
That part kind of confuses me.
AgNO3(aq) + Cl^-(aq) ==> AgCl(s) + NO3^-(aq)
Thank you!
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?
To determine the mass of silver nitrate required to completely precipitate the chloride ions from the sample, you need to follow a few steps:
Step 1: Calculate the mass of chloride ion in the 1.054 g sample.
Given that the sample is known to contain 10.3% chloride ion by mass, you can calculate the mass of chloride ion as follows:
Mass of chloride ion = 10.3% of 1.054 g
Step 2: Convert the mass of chloride ion to moles.
To convert the mass of chloride to moles, you need to use the molar mass of chloride ion. The molar mass of chloride is 35.45 g/mol.
Step 3: Use the stoichiometry of the balanced equation to determine the molar ratio between chloride ion and silver nitrate.
From the balanced equation:
1 mole of AgCl is formed from 1 mole of Ag+ and 1 mole of Cl-.
Step 4: Use the molar ratio to calculate the moles of silver nitrate required.
Since the molarity ratio between Ag+ and Cl- is 1:1, the moles of Ag+ will be equal to the moles of Cl-.
Step 5: Convert the moles of AgNO3 to grams.
To convert moles of silver nitrate to grams, use the molar mass of AgNO3, which is 169.87 g/mol.
Step 6: Determine the mass of silver chloride obtained.
Since the stoichiometry of the balanced equation shows a 1:1 molar ratio between AgCl and Cl-, the mass of AgCl obtained will be equal to the mass of chloride ion in the sample.
By following these steps, you will determine the mass of silver nitrate needed to precipitate the chloride ions and the mass of silver chloride produced in the reaction.
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.