A precipitation reaction occurs. Yellow precipitates of lead iodide are formed. The equation for this reaction is
Pb(NO3)2 + 2KI -----> PbI2 + 2KNO3
When potassium iodide reacts with lead nitrate, a double displacement reaction occurs. The potassium ion and the lead ion switch places to form potassium nitrate and lead iodide. This reaction results in the formation of a yellow precipitate of lead iodide.
When potassium iodide and lead nitrate are combined, a double displacement reaction occurs, resulting in the formation of lead iodide and potassium nitrate. Lead iodide is a yellow precipitate that forms when the two solutions are mixed.
When potassium iodide reacts with barium nitrate, a double displacement reaction occurs. The potassium ions and barium ions switch places to form potassium nitrate and barium iodide. Both products are insoluble and will form a precipitate.
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.
The reaction between silver nitrate and potassium iodide forms silver iodide precipitate and potassium nitrate. This reaction is a double displacement reaction where the silver ions from silver nitrate switch places with the potassium ions in potassium iodide.
The products are Mercury(II) iodide and Potassium nitrate
When potassium iodide reacts with lead nitrate, a double displacement reaction occurs. The potassium ion and the lead ion switch places to form potassium nitrate and lead iodide. This reaction results in the formation of a yellow precipitate of lead iodide.
Silver iodide (AgI), a precipitate insoluble in water, don't react with potassium nitrate.
When potassium iodide and lead nitrate are combined, a double displacement reaction occurs, resulting in the formation of lead iodide and potassium nitrate. Lead iodide is a yellow precipitate that forms when the two solutions are mixed.
When potassium iodide reacts with barium nitrate, a double displacement reaction occurs. The potassium ions and barium ions switch places to form potassium nitrate and barium iodide. Both products are insoluble and will form a precipitate.
A yellow precipitate of lead iodide is formed due to the reaction between potassium iodide and lead nitrate. This reaction is a double displacement reaction, where the potassium from potassium iodide swaps places with the lead from lead nitrate, forming the insoluble lead iodide.
A white precipitate of silver iodide forms due to the reaction between silver ions and iodide ions, leaving potassium nitrate in solution. This reaction is a double displacement reaction and is used as a test for iodide ions.
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.
In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.
The reaction between silver nitrate and potassium iodide forms silver iodide precipitate and potassium nitrate. This reaction is a double displacement reaction where the silver ions from silver nitrate switch places with the potassium ions in potassium iodide.
Silver nitrate solution and potassium iodide solution can be mixed to form silver iodide due to a double displacement reaction where the silver ions from silver nitrate react with the iodide ions from potassium iodide to form insoluble silver iodide precipitate. The balanced chemical equation for this reaction is: AgNO3 (aq) + KI (aq) → AgI (s) + KNO3 (aq).
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3