Aqueous lead nitrate plus aqueous sodium iodide produce solid lead iodide and aqueous sodium nitrate.
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
Potassium iodide + silver nitrate --> Silver iodide and potassium nitrate The chemical equation is: K+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + K+[NO3]- (aq)
Pb(NO3)2(aq)+2NaI(aq)=2NaNO3(aq)+PbI2(s)
Ag(NO3)(aq) + KI(aq) ---> K(NO3)(aq) + AgI(s)
A yellow precipitate of silver iodide (AgI).
A precipitate of Lead iodide and Potassium nitrate are formed
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
Yes. The equation is Pb(NO3)2(aq) + 2KI(aq) ==> PbI2(s) + 2 KNO3(aq)
Pour a solution of Sodium(or Potassium) Iodide over Lead nitrate solution. The Lead iodide will be precipitated out as a yellow solid
Chloride and iodide ions can be distinguished by the colour of their precipitate which are formed by treatig it with silver nitrate solution.
Potassium iodide + silver nitrate --> Silver iodide and potassium nitrate The chemical equation is: K+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + K+[NO3]- (aq)
The net ionic equation is Ag+(aq) + I- --> AgI(s) The sodium (Na+) and nitrate (NO3-) are spectators as always.
The chemical equation is:2 KI + Pb(NO3)2 = 2 KNO3 + PbI2(s)
Pb(NO3)2(aq)+2NaI(aq)=2NaNO3(aq)+PbI2(s)
A precipitate of yellow Lead iodide and Sodium nitrate are formed