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

Answer:

Well, #"w/v"-="Mass of solute"/"Volume of solution"#, we should get over #7*g# of silver chloride.

Explanation:

And thus #"mass of silver nitrate"=16.9%xx50*mL=8.45*g#.

And this represents a molar quantity of #(8.45*g)/(169.87*g*mol^-1)#

#=0.0497*mol#, with respect to #AgNO_3#.

Likewise #"mass of sodium chloride"=5.8%xx50*mL=2.90*g#.

And this represents a molar quantity of #(2.90*g)/(58.44*g*mol^-1)#

#=0.0497*mol# 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^(+) + Cl^(-)rarrAgCl(s)darr#

Of course the complete reaction is....

#AgNO_3(aq) + NaCl(aq) rarr AgCl(s)darr + NaNO_3(aq)#, i.e. sodium nitrate remains in solution and can be separated (with effort) from the precipitate.

And given the stoichiometry, we gets #0.04974*molxx143.32*g*mol^-1=7.11*g#.

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.