Here is the equation:
Cl2(aq) + 2 KI(aq) ----> 2 KCl(aq) + I2(aq)
single-replacement
potassium nitrate would be left was an aqueous solution and lead iodide would be the precipitate
NO, they are different.Iodide is only one ion (I-) and potassium iodide (KI) is the salty product when you react potassium (K) and iodine (I2)
Displacement scratch that it a single replacement
It is left as a low hazard
Yes, it is correct.
2KI + Cl2 = 2KCl + I2
Chlorine displaces Potassium Iodide to liberate aqueous I2(brown colour). Hence the solution turns brown.
This is the correct answer: Cl2(g)+2KI(aq) = I2(s)+2KCl(aq)
single-replacement
Potassium chloride and Iodine
Chlorine, a more reactive halogen would displace iodide in its hallide solution. Potassium chloride would be formed.
potassium nitrate would be left was an aqueous solution and lead iodide would be the precipitate
NO, they are different.Iodide is only one ion (I-) and potassium iodide (KI) is the salty product when you react potassium (K) and iodine (I2)
It is the Iodine dissolved in aqueous Potassium(or Sodium) Iodide
When chlorine gas is bubbled into an aqueous solution of potassium iodide, some of the iodide ions are oxidized to iodine. The iodine molecules combine with iodide ions to form brown triiodide ion, I3-. In this demonstration, the aqueous solution is above a layer of carbon tetrachloride, in which iodine is quite soluble. The beautiful violet color of iodine can be seen as the iodine dissolves in the carbon tetrachloride layer. With excess chlorine, iodine reacts to form iodine monochloride, ICl, which is ruby red. The iodine monochloride reacts further to form iodine trichloride, ICl3, which is much lighter in color, causing the solution to be decolorized.
In aqueous solution they would not react. They would form a solution of ferric ions, chloride ions, potassium ions, and iodide ions.