Lead iodide (Pb2I) precipitates as a yellow solid, leaving a solution of potassium and nitrate ions.
Ag(NO3)(aq) + KI(aq) ---> K(NO3)(aq) + AgI(s)
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
A yellow precipitate of silver iodide (AgI).
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.
In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.
Ag(NO3)(aq) + KI(aq) ---> K(NO3)(aq) + AgI(s)
A precipitate of Lead iodide and Potassium nitrate are formed
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
It produces Potassium nitrate and Lead iodide
no
potassium nitrate would be left was an aqueous solution and lead iodide would be the precipitate
Pour a solution of Sodium(or Potassium) Iodide over Lead nitrate solution. The Lead iodide will be precipitated out as a yellow solid
Silver iodide (AgI), a precipitate insoluble in water, don't react with potassium nitrate.
A yellow precipitate of silver iodide (AgI).
This is a double displacement reaction. 2KI + Pb(NO3)2 --> 2KNO3 + PbI2 Potassium iodide + Lead(II) nitrate --> Potassium nitrate + Lead(II) iodide A bright yellow precipitate will form when these two react.
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.
nothing