What is the mass of the precipitate formed when 50 mililitres of 16.9%(w/v)solution of silver nitrate is mixed with 50 mililitres of 5.8%(w/v) sodium chloride solution?
Answers
Answered by
1
Well, w/v≡Mass of solute/Volume of solution
we should get over 7⋅g of silver chloride.
And thus mass of silver nitrate=16.9%×50⋅mL=8.45⋅g.
And this represents a molar quantity of 8.45⋅g169.87⋅g⋅mol−1
=0.0497⋅mol, with respect to AgNO3.
Likewise mass of sodium chloride=5.8%×50⋅mL=2.90⋅g.
And this represents a molar quantity of 2.90⋅g58.44⋅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−→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 the stoichiometry, 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 moreover AgCl is photoactive, and would decompose under light.
we should get over 7⋅g of silver chloride.
And thus mass of silver nitrate=16.9%×50⋅mL=8.45⋅g.
And this represents a molar quantity of 8.45⋅g169.87⋅g⋅mol−1
=0.0497⋅mol, with respect to AgNO3.
Likewise mass of sodium chloride=5.8%×50⋅mL=2.90⋅g.
And this represents a molar quantity of 2.90⋅g58.44⋅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−→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 the stoichiometry, 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 moreover AgCl is photoactive, and would decompose under light.
Similar questions