Produces yellow Lead(II) iodide and Sodium nitrate
Pb(NO3)2 + 2NaI = PbI2 = 2NaNO3
Lead iodide is a yellow precipitate.
Aqueous lead nitrate plus aqueous sodium iodide produce solid lead iodide and aqueous sodium 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.
Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq) Aqueous lead II nitrate reacts with aqueous sodium iodide to form solid lead II iodide precipitate and aqueous sodium nitrate.
In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.
Pb(NO3)2(aq)+2NaI(aq)=2NaNO3(aq)+PbI2(s)
Lead(II) iodide and Sodium nitrate
A precipitate of yellow Lead iodide and Sodium nitrate are formed
It produces Potassium nitrate and Lead iodide
Aqueous lead nitrate plus aqueous sodium iodide produce solid lead iodide and aqueous sodium nitrate.
Pour a solution of Sodium(or Potassium) Iodide over Lead nitrate solution. The Lead iodide will be precipitated out as a yellow solid
Na2
A yellow Lead(II) iodide precipitate
The lead nitrate and sodium sulfate precipitate together and becomes lead sulfate and sodium nitrate. lead nitrate+ sodium sulfate --> lead sulfate + sodium 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.
Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq) Aqueous lead II nitrate reacts with aqueous sodium iodide to form solid lead II iodide precipitate and aqueous sodium nitrate.
Take a few drops of both samples and add some lead nitrate. A yellow precipitate indicates lead iodide and it gives the inference that it contains iodide ions, hence the solution of sodium iodide.
A double displacement reaction.