A solution contains an unknown mass of dissolved silver ions. When potassium chloride is added to the solution, a white precipitate forms. The precipitate is filtered and dried and found to have a mass of 212 mg. What mass of silver was in the original solution? (Assume that all of the silver was precipitated out of solution by the reaction.)
Amount of precipitate formed = 212 mg
The chemical equation for the reaction happening here is :
Ag2+ (aq) + KCl (aq) → AgCl (s) + K+
Let’s find out the number of moles of AgCl formed :
Number of moles (n) =
Molar mass of AgCl = 143.32 g/mol.
Barium precipitate formed is in mg, let’s convert to g :
1 g = 1000 mg
Therefore, 0.212 mg in gram is :
212 mg= 0.212 g
Now, let’s find the number of moles :
n = 0.0014 moles
Therefore, the number of moles of Ag+ in the original solution is 0.0014 moles.