KOH and NaOH with Cl2?

You must think of this as occurring in two steps. First step is the reaction of the Cl2 with the NaOH: NaOH + Cl2? NaOCl + HCl The HCl that is produced reacts with the NaOH as follows: NaOH + HCl?

NaCl + H2O If you add these two equations together and cancel out what is common to both sides , you get: 2NaOH + Cl2? NaOCl + NaCl + H2O This is the equation that you found on the internet - this represents what happens. You attempt shows only the reaction of Cl 2 with the water: Cl2 + H2O?

HOCl + HCl But this does not include any reaction with the NaOH - which is what your question requires So now you proceed: HOCl + HCl + 2NaOH? NaOCl + NaCl + 2H2O Now once again add these two equations together - delete anything common: Cl2 + 2NaOH? NaOCl + NaCl + H2O Which brings us back to the same conclusion we reached above.

Note: NaOCl and NaClO , and also HOCl and HClO, are the same compounds - just different ways of writing the formula. I do not think that your statement: the equation in my book is (Cl2 + NaOH = HOCl + Cl - + H+) is correct - this equation is not correct - what happened to the Na on the products side?

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