Dissolve each of the silver nitrate and potassium iodide separately in water, then mix the two solutions slowly with stirring. Silver iodide will precipitate and can be separated by filtering it from the liquid.
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
Potassium nitrate and a precipitate of Silver iodide are formed
It produces Potassium nitrate and Lead iodide
A precipitate of Lead iodide and Potassium nitrate are formed
potassium nitrate would be left was an aqueous solution and lead iodide would be the precipitate
The products are Mercury(II) iodide and Potassium nitrate
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
Potassium nitrate and a precipitate of Silver iodide are formed
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
It produces Potassium nitrate and Lead iodide
In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.
A precipitate of Lead iodide and Potassium nitrate are formed
Ag(NO3)(aq) + KI(aq) ---> K(NO3)(aq) + AgI(s)
Potassium iodide + silver nitrate --> Silver iodide and potassium nitrate The chemical equation is: K+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + K+[NO3]- (aq)
Silver iodide (AgI), a precipitate insoluble in water, don't react with potassium nitrate.
2KI+Pb(NO(3))(2) yields 2KNO(3)+PbI(2). You basically get potassium nitrate and lead (II) iodide when you react potassium iodide and lead nitrate dissolved in solution.
potassium nitrate would be left was an aqueous solution and lead iodide would be the precipitate