in aqueous medium ppts. of copper iodide are formed which are converted into cuprous iodide and free iodine in a short time. 2NaI + Cu(NO3)2 = CuI2 + 2NaNO3 2CuI2 = Cu2I2 + I2
Aqueous lead nitrate plus aqueous sodium iodide produce solid lead iodide and aqueous sodium nitrate.
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.
Produces Silver iodide precipitate and Sodium nitrate
Equation: NaI + AgNO3 ----> NaNO3 + AgI
Produces yellow Lead(II) iodide and Sodium nitrate
the equation for sodium nitrate and copper III iodide can be given below.Cu I 2 +2 Na No3 ->I (NO3)2 + 2NaI. this is the balance reaction for the above.
Aqueous lead nitrate plus aqueous sodium iodide produce solid lead iodide and aqueous sodium nitrate.
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.
The equation for the reaction between silver nitrate and sodium iodide is AgNO3 + NaI -> AgBr + NaNO3. The gram formula masses are 169.87 for silver nitrate, 149.89 for sodium iodide, and 84.99 for sodium nitrate. Therefore, 1.7 g of silver nitrate constitutes 1.7/169.87 or 0.010 formula mass of silver nitrate and 1.5 g of sodium iodide constitutes 1.5/149.89 or 0.010 mole of sodium iodide, to the justified number of significant digits. The reaction equation shows that the number of formula unit masses of each reactant and product are the same, so that there will be 0.85 g of sodium nitrate produced, to the justified number of significant digits.
Produces Silver iodide precipitate and Sodium nitrate
Lead(II) nitrate and sodium iodide will yield lead(II) iodide and sodium nitrate. This is a double displacement reaction, where the cations and anions switch partners resulting in the formation of two new compounds.
Equation: NaI + AgNO3 ----> NaNO3 + AgI
The net ionic equation for the reaction between sodium iodide (NaI) and silver nitrate (AgNO3) when a precipitate is formed is: 2Ag+ + 2I- -> Ag2I (s) This equation represents the formation of silver iodide (AgI) precipitate when silver cations react with iodide anions.
When lead nitrate is mixed with sodium iodide, a solid precipitate of lead iodide is formed along with sodium nitrate. This reaction is a double displacement reaction where the cations of the two compounds switch partners to form the products. Lead iodide is a yellow precipitate that can be easily observed in the reaction mixture.
Produces yellow Lead(II) iodide and Sodium nitrate
Na2S + Cu(NO3)2 -> 2NaNO3 + CuS
The chemical equation is: Na+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + Na+[NO3]- (aq)