Silver ions can be precipitated from aqueous solutions by the addition of aqueous chloride:
Ag+(aq) + Cl-(aq) → AgCl(s)
Silver chloride is virtually insoluble in water so that the reaction appears to go to completion. How many grams of solid NaCl must be added to 25.0 mL of 0.366 M AgNO3 solution to completely precipitate the silver? Please show the work
Ag+(aq) + Cl-(aq) → AgCl(s)
Silver chloride is virtually insoluble in water so that the reaction appears to go to completion. How many grams of solid NaCl must be added to 25.0 mL of 0.366 M AgNO3 solution to completely precipitate the silver? Please show the work
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Answer
AgNO3+ NaCl ----> AgCl + NaNO3
Ag+(aq) + Cl-(aq) → AgCl(s)
1 mol AgNO3 and 1 mol NaCl combine to make 1 mol AgCl
Molar mass : NaCl, 58.5 g / mol, AgNo3, 169.9 g / mol
25 ml 0.366 M AgNO3 is 0.366*25/1000= 0.00915 moles of AgNO3
0.00915 moles of AgNO3 would require same number of moles of NaCl, mass of NaCl required is therefore 0.00915*58.5 = 0.5353 g
AgNO3+ NaCl ----> AgCl + NaNO3
Ag+(aq) + Cl-(aq) → AgCl(s)
1 mol AgNO3 and 1 mol NaCl combine to make 1 mol AgCl
Molar mass : NaCl, 58.5 g / mol, AgNo3, 169.9 g / mol
25 ml 0.366 M AgNO3 is 0.366*25/1000= 0.00915 moles of AgNO3
0.00915 moles of AgNO3 would require same number of moles of NaCl, mass of NaCl required is therefore 0.00915*58.5 = 0.5353 g