Part III: TRIAL 1 TRIAL 2
Mass of sample KNO3 1.039g 1.041g
Drops of AgNO3 27 drops 28 drops
1. Net ionic eq. for formation of solid silver chloride from aqueous solution
NaCl(aq)+AgNO3(aq)→ NaNO3(aq)+AgCl(s)
Na+(aq)+Cl-(aq)+Ag+(aq)+NO3-(aq)→ Na+(aq)+NO3-(aq)+AgCl(s) Cl-(aq)+Ag+(aq)→ AgCl(s)
2. Net ionic eq. for formation of solid silver chromate from aqueous solution
K2CrO4(aq)+2AgNO3(aq)→ 2KNO3(aq)+Ag2CrO4(s)
2K+(aq)+CrO42-(aq)+2Ag+(aq)+2NO3-(aq)→ 2K+(aq)+2NO3-(aq)+Ag2CrO4(s) 2Ag+(aq)+CrO42-(aq)→ Ag2CrO4(s)
3. Calculation of mass percent NaCl in sample of crude KNO3 from mass of NaCl and mass of sample
Average of drops: (27+28)/2 = 27.5 drops
27.5 drops x (1mL/18.78 drops) x (0.5mol AgNO3/1L soln) x (1L soln/1000mL soln) …show more content…
The other three possible salts were not so much. By reducing the temperature, the KNO3 crystals began to form, leaving behind the other three, which are too soluble to even crystallize under such low temperature. In part II, the filtrate now contains more chloride and sodium ions, so NaCl is to be removed next. Since NaCl is the least soluble out of the four possible salts at the majority of temperatures, but does not change solubility very much with the change of temperature. Therefore, the best way to crystallize NaCl is by evaporation. Through evaporation, the concentrations of sodium ions and chloride ions increased. The concentrations of the ions were so high that not all the ions can remain dissolved, therefore forming crystals of NaCl. The more the filtrate evaporates on the hot plate, the more NaCl crystals began to form. The theoretical yield is what would be if 100% of the reaction was complete and no amount of products were lost. But this all depends the technical perfection of doing the experiment. Reactions can sometimes be reversible; therefore less than 100% of the reaction goes to completion. Some other reactions require very high amount of energy or extended time so it cannot go to
Mass of sample KNO3 1.039g 1.041g
Drops of AgNO3 27 drops 28 drops
1. Net ionic eq. for formation of solid silver chloride from aqueous solution
NaCl(aq)+AgNO3(aq)→ NaNO3(aq)+AgCl(s)
Na+(aq)+Cl-(aq)+Ag+(aq)+NO3-(aq)→ Na+(aq)+NO3-(aq)+AgCl(s) Cl-(aq)+Ag+(aq)→ AgCl(s)
2. Net ionic eq. for formation of solid silver chromate from aqueous solution
K2CrO4(aq)+2AgNO3(aq)→ 2KNO3(aq)+Ag2CrO4(s)
2K+(aq)+CrO42-(aq)+2Ag+(aq)+2NO3-(aq)→ 2K+(aq)+2NO3-(aq)+Ag2CrO4(s) 2Ag+(aq)+CrO42-(aq)→ Ag2CrO4(s)
3. Calculation of mass percent NaCl in sample of crude KNO3 from mass of NaCl and mass of sample
Average of drops: (27+28)/2 = 27.5 drops
27.5 drops x (1mL/18.78 drops) x (0.5mol AgNO3/1L soln) x (1L soln/1000mL soln) …show more content…
The other three possible salts were not so much. By reducing the temperature, the KNO3 crystals began to form, leaving behind the other three, which are too soluble to even crystallize under such low temperature. In part II, the filtrate now contains more chloride and sodium ions, so NaCl is to be removed next. Since NaCl is the least soluble out of the four possible salts at the majority of temperatures, but does not change solubility very much with the change of temperature. Therefore, the best way to crystallize NaCl is by evaporation. Through evaporation, the concentrations of sodium ions and chloride ions increased. The concentrations of the ions were so high that not all the ions can remain dissolved, therefore forming crystals of NaCl. The more the filtrate evaporates on the hot plate, the more NaCl crystals began to form. The theoretical yield is what would be if 100% of the reaction was complete and no amount of products were lost. But this all depends the technical perfection of doing the experiment. Reactions can sometimes be reversible; therefore less than 100% of the reaction goes to completion. Some other reactions require very high amount of energy or extended time so it cannot go to