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What mass of silver nitrate must be used to completely precipitate the chloride ion from the sample?

Dec 23, 2008 by angela | Posted in Chemistry

please mitigate me out asap!!!

Suppose a 1.441-g sample is known to contain 12.5 % chloride ion by mass. What mass of silver nitrate must be old to completely precipitate the chloride ion from the sample? What mass of silver chloride will be obtained?

Hi angela....im stuck on the same uncontrollable...hahahaha

Silver nitrate is highly soluble salt and when it is dissolved it forms silver and nitrate ions:
AgNO3(aq) -----> Ag^+(aq) + NO3^-(aq)

Silver ions get even with chloride ions present in the solution and forms insoluble silver chloride:

Ag^+(aq) + Cl^-(aq) -------> AgCl(s)
This is the net ionic equation of the retaliation.

Mass of the sample:1.441 g
Mass of Cl- ions in the sample: 1.441 x (12.5/100) = 0.180125 g
Molar mass of Cl = 35.5 g /mol
Mole of Cl- ions = 0.180125 g / 35.5 g/mol = 0.0050739mol

According to net ionic equation;
1 mol Ag^+ ion is inexorable to precipitate 1 mol Cl^- ion,
therefore for 0.0050739 mol Cl^- ion we will need 0.0050739 mol Ag^+ ion.

According to dissolving equation;
AgNO3(aq) -----> Ag^+(aq) + NO3^-(aq)

0.0050739 mol Ag^+ ion will be produced by 0.0050739 mol AgNO3

Molar mass of AgNO3 = 170 g /mol
Mass of AgNO3 required : 0.0050739 mol x 170 g/mol = .863 g

What mass of silver nitrate is required to is required to react with this amount of sodium chloride?

Aug 27, 2008 by Ravz | Posted in Chemistry

Silver nitrate reacts with sodium chloride in a 1:1 relationship, as given in the following equation.
AgNO + NaCl --> AgCl + NaNO

The label on a bottled not function claims to contain 35 mg (0.035 g) of sodium chloride. what mass of silver nitrate is required to react with this amount of sodium chloride?

1) Silver nitrate is AgNO3 and not AgNO. Your advisor doesn't like that.

2) You need one mol NaCl for the reactiom with one mol AgNO3

n= 1= m(AgNO3)/M(AgNO3) = m(NaCl)/M(NaCl) (1)

m: mass and M : molar mass

M(AgNO3)= 170 g/mol and M(NaCl)= 58 g/mol
with m(NaCl)= 35 mg and rearranging equation 1 one gets:

m(AgNO3)= 102 mg

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