Well,w/v≡Mass of soluteVolume of solution, we should get over7⋅gof silver chloride.

Explanation:

And thusmass of silver nitrate=16.9%×50⋅mL=8.45⋅g.

And this represents a molar quantity of8.45⋅g/169.87⋅g⋅mol−1

=0.0497⋅mol, with respect toAgNO3.

Likewisemass of sodium chloride=5.8%×50⋅mL=2.90⋅g.

And this represents a molar quantity of2.90⋅g/58.44⋅g⋅mol−1

=0.0497⋅molwith respect toNaCl.

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

Ag++Cl−→AgCl(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 thestoichiometry, we gets,

0.04974⋅mol×143.32⋅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 moreoverAgClis photoactive, and would decompose under light.