530,3 g potassium iodide are needed.
Lead iodide can be obtained as a yellow precipitate by reacting solutions oflead(II) nitrate and potassium iodide the other compound created by the reaction is potassium nitrate which is a white power but as it is soluble it does not form a precipitate. 2KI + Pb(NO3)2 --> 2KNO3 + PbI2 the balanced equation if you were needing it
is it potassium iodide
The boiling point of potassium iodide is 1 330 0C. The boiling point of potassium chloride is 1 420 0C.
The only iron iodide listed in the Handbook of Chemistry and Physics (1985) is iron (II) iodide, with the formula FeI2 Hypothetically, there could be an iron (III) iodide with the formula FeI3, but this probably spontaneously transforms into iron (II) iodide and elemental iodine.
Chromium (III) Iodide
Potassium iodide is added in excess to ensure that all available lead nitrate has reacted to form lead iodide. This helps to maximize the yield of lead iodide and ensures that there is no excess lead nitrate remaining in the solution.
The products are Mercury(II) iodide and Potassium nitrate
In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.
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.
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 + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
potassium nitrate would be left was an aqueous solution and lead iodide would be the precipitate
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.
Silver iodide (AgI), a precipitate insoluble in water, don't react with potassium nitrate.
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.
The balanced equation is 2 KI + Pb(NO3)2 -> 2 KNO3 + PbI2.
Potassium Iodide- KI Lead Nitrate- Pb(NO3)2