What is the mass of precipitate formed when 50 ml of 16.9% w/v solution of AgNO3 is mixed with 50 ml of 5.8% w/v NaCl solution?

1 Answer
Aug 17, 2017

Well, w/vMass of soluteVolume of solution, we should get over 7g of silver chloride.

Explanation:

And thus mass of silver nitrate=16.9%×50mL=8.45g.

And this represents a molar quantity of 8.45g169.87gmol1

=0.0497mol, with respect to AgNO3.

Likewise mass of sodium chloride=5.8%×50mL=2.90g.

And this represents a molar quantity of 2.90g58.44gmol1

=0.0497mol with respect to NaCl.

Clearly the reagents are present in 1:1 molar ratio. The reaction that occurs in solution is the precipitation of a curdy white mass of AgCl(s), i.e. and we represent this reaction by the net ionic equation.....

Ag++ClAgCl(s)

Of course the complete reaction is....

AgNO3(aq)+NaCl(aq)AgCl(s)+NaNO3(aq), i.e. sodium nitrate remains in solution and can be separated (with effort) from the precipitate.

And given the stoichiometry, we gets 0.04974mol×143.32gmol1=7.11g.

Of course a material such as silver halide would be very hard to isolate. Particle size is very small; it is likely to clog the filter, and filter very slowly; and moreover AgCl is photoactive, and would decompose under light.