This is the correct answer:
Cl2(g)+2KI(aq) = I2(s)+2KCl(aq)
2KI + Cl2 = 2KCl + I2
Potassium chloride and Iodine
Chlorine, a more reactive halogen would displace iodide in its hallide solution. Potassium chloride would be formed.
In aqueous solution they would not react. They would form a solution of ferric ions, chloride ions, potassium ions, and iodide ions.
Chlorine displaces Potassium Iodide to liberate aqueous I2(brown colour). Hence the solution turns brown.
Yes, it is correct.
2KI + Cl2 = 2KCl + I2
Potassium chloride and Iodine
Chlorine, a more reactive halogen would displace iodide in its hallide solution. Potassium chloride would be formed.
In aqueous solution they would not react. They would form a solution of ferric ions, chloride ions, potassium ions, and iodide ions.
Worded Equation; Potassium Iodide + Calcium Chloride ------> Potassium Chloride + Calcium Iodide Chemical Equation; KI (l) + CaCl (l) -----> KCl (aq) + Ca(I)2 (aq) Note Answer is only correct if proper states are applied and used.
Chlorine displaces Potassium Iodide to liberate aqueous I2(brown colour). Hence the solution turns brown.
No - there would be a reaction though if Chlorine and Potassium Iodide were mixed
Displacement scratch that it a single replacement
Here is the equation:Cl2(aq) + 2 KI(aq) ----> 2 KCl(aq) + I2(aq)
you have to write... 2KI + Cl2 = 2KCl + I2
2KI + Cl2 = 2KCl + I2