What would the product be:
1.NaBr + CI2
2.NaBr + Br2
Would the products be:
1. Iodine?
2. Bromine?
Thank You!
3 answers
1,2 dichloroethane was also added to 1 and 2
Woah major brain fart. Would the products be:
1.2NaCl+I2?
2.Na2 + Br2?
1.2NaCl+I2?
2.Na2 + Br2?
You can't possibly get I2 out of a solution of NaBr and Cl2.
Cl2 is a good oxidizing agent and will oxidize Br^- (which you have in NaBr) to Br2. Adding a solvent as you did, the Br2 dissolves in that and give an amber to orange color depending upon how much Br2 is produced.
2NaBr + Cl2 ==> 2NaCl+ Br2
NaBr + Br2 makes no sense. There is no reaction.
These usually are done in connection with identification of iodides (such as NaI, KI, etc) and bromides (such as NaBr, KBr, etc).
Chlorine (Cl2) oxidizes both I^- and Br^- to the free element. Adding an organic solvent (I always use CCl4 but that's a no no in labs now so they use chloroform, hexane, or some such solvent). Free I2 turns violet and free Br2 turns amber to orange depending upon the amount of Br2. The equationsare
2NaI + Cl2 ==> 2NaCl+ I2
2NaBr + Cl2 ==> 2NaCl + Br2
Cl2 is a good oxidizing agent and will oxidize Br^- (which you have in NaBr) to Br2. Adding a solvent as you did, the Br2 dissolves in that and give an amber to orange color depending upon how much Br2 is produced.
2NaBr + Cl2 ==> 2NaCl+ Br2
NaBr + Br2 makes no sense. There is no reaction.
These usually are done in connection with identification of iodides (such as NaI, KI, etc) and bromides (such as NaBr, KBr, etc).
Chlorine (Cl2) oxidizes both I^- and Br^- to the free element. Adding an organic solvent (I always use CCl4 but that's a no no in labs now so they use chloroform, hexane, or some such solvent). Free I2 turns violet and free Br2 turns amber to orange depending upon the amount of Br2. The equationsare
2NaI + Cl2 ==> 2NaCl+ I2
2NaBr + Cl2 ==> 2NaCl + Br2
1 answer