lead iodide crystals.. check this site out for the reaction answer.. http://en.wikipedia.org/wiki/Lead(II)_iodide
In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.
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.
When potassium iodide is added to lead nitrate, a precipitation reaction occurs resulting in the formation of lead iodide, a yellow insoluble solid, and potassium nitrate, which remains in solution. This reaction can be visually identified by the formation of a yellow precipitate.
Potassium iodide + silver nitrate --> Silver iodide and potassium nitrate The chemical equation is: K+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + K+[NO3]- (aq)
Aqueous lead nitrate plus aqueous sodium iodide produce solid lead iodide and aqueous sodium nitrate.
You can separate lead iodide precipitate and potassium nitrate solution by filtration. Pour the mixture through a filter paper in a funnel to trap the solid lead iodide, allowing the clear potassium nitrate solution to pass through. The lead iodide will be left on the filter paper, which can then be washed with water to remove any remaining potassium nitrate before drying the solid lead iodide.
In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.
When silver nitrate reacts with potassium iodide, a precipitation reaction occurs where silver iodide is formed. The balanced chemical equation for this reaction is: AgNO3 + KI -> AgI + KNO3. The silver iodide formed will appear as a yellow solid 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.
When potassium iodide is added to lead nitrate, a precipitation reaction occurs resulting in the formation of lead iodide, a yellow insoluble solid, and potassium nitrate, which remains in solution. This reaction can be visually identified by the formation of a yellow precipitate.
Potassium iodide + silver nitrate --> Silver iodide and potassium nitrate The chemical equation is: K+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + K+[NO3]- (aq)
AgNO3(aq) + KI(aq) = KNO3(aq) + AgI(s) This is a classic test for halogens, and AgI precipitates down as a yellow solid.
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 that occurs is a double displacement reaction where lead(II) nitrate and potassium iodide switch partners to form solid lead(II) iodide and potassium nitrate solution. This reaction can be represented by the equation Pb(NO3)2 + 2KI -> Pbl2 + 2KNO3.
Lead(II) nitrate (Pb(NO3)2) and potassium iodide (KI) are ionic compounds that are typically found as solids at room temperature. When mixed together in an aqueous solution, they can form a precipitate of lead(II) iodide (PbI2), which is also a solid.
Lead iodide is prepared from lead nitrate because lead nitrate contains lead ions and nitrate ions that can react with iodide ions to form lead iodide. This reaction allows for the precipitation of lead iodide, which can then be isolated and collected.
They form Silver Iodide and Poassium nitrate. Silver Iodide is a yellow coloured compound which forms precipitate in the solution. Potassium Nitrate remains dissolved. These reactions only take place in solution. When both reactants are in solid state then reaction may not occur, or it is too slow to be observed.