The 2 steps to the reaction between NaX2COX3NaX2COX3 and HClHCl are:
If NaX2COX3NaX2COX3 is in excess and we add limited amount of HClHCl (e.g. 2:1), why is it that after the addition of limited HClHCl , we still have NaHCOX3NaHCOX3 present together with NaX2COX3NaX2COX3?
What I was thinking is that as NaHCOX3NaHCOX3 is formed, the HClHCl in the solution will react with it through step 2 to form COX2COX2 and HX2OHX2O. So, half of the HClHCl will react with NaX2COX3NaX2COX3 and the other half will react with NaHCOX3NaHCOX3. And so, if this happens there will only be NaX2COX3NaX2COX3 present at the end. (since all the NaHCOX3NaHCOX3 that is formed from step 1 will be reacted with HClHCl upon formation)
Why wouldn't this be the case? Why must step 2 only occur after step 1 is completed?
