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In World War 1, the chlorine gas was handled by some soldiers by damping their hankerchiefs into sodium carbonate solution. The article chlorine monoxide suggests that the reaction release chlorine monoxide at 20-30 degree Celcius. I wonder how is the gas released at this temperature since the concentration of sodium carbonate was not given.

From this paper, it says:

Under the optimal conditions of reaction, sodium carbonate, containing approximately 10% of water, is allowed to fall slowly down a reaction tower and contacted with an upward flow of chlorine diluted to about 25% in a gas such as air, nitrogen, or carbon dioxide. The humidity in the reaction zone must be maintained at around 10%, either by saturating the incoming gas with water or by injecting steam into the reactor at regular intervals. The temperature of the reaction zone is preferably maintained at about room temperature (20-30 ᵒC), but may be raised to as high as 200 ᵒC.

So, I figure that under small amount of water, chlorine gas could form HClO via

Cl2 + H2O -> HCl + HClO

My question is:

  1. What is the mechanism of Cl2O formation?
  2. How does the concentration of sodium carbonate and temperature would affect the formation of Cl2O compound?
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1 Answer 1

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$\ce{Cl2O}$ is simply formed by dehydration of $\ce{HClO}$ according to : $$\ce{2 HClO -> Cl2O + H2O}$$ $\ce{Cl2O}$ can be considered as an anhydride of the acid $\ce{HClO}$.

$\ce{Na2CO3}$ is used to push the reaction of chlorine on water to the right-hand-side, as $\ce{Na2CO3}$ destroys $\ce{HCl}$ as soon as produced, which prevents back-reaction to occur. The equations are : $$\ce{Cl2 + H2O <=> HCl + HClO}$$ $$\ce{Na2CO3 + 2 HCl -> 2 NaCl + H2O + CO2}$$

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  • $\begingroup$ A simple and good answer. But, assuming the excess of carbonate, I would rather write $\ce{Na2CO3 + HCl -> NaHCO3 + NaCl}$. $\endgroup$
    – Poutnik
    Feb 15 at 8:26
  • $\begingroup$ @Poutnik.OK. Thank you. I know and I have hesitated to show "your" reaction, with the formation of $\ce{NaHCO3}$ $\endgroup$
    – Maurice
    Feb 15 at 10:28
  • $\begingroup$ Hehe, I have hesitated to edit your answer, I let it often on the question author if he/wants to use it. $\endgroup$
    – Poutnik
    Feb 15 at 13:19
  • $\begingroup$ So, regardless of the concentration of Na2CO3, chlorine monoxide would always be produced, wouldn't it? $\endgroup$
    – Seiji
    Feb 15 at 15:18
  • $\begingroup$ @Seiji. I am not a specialist of this chemistry. Apparently the concentration of $\ce{Na2CO3}$ is not crucial. Its total amount is more important. It must be sufficient to destroy all $\ce{HCl}$ produced. But for example, I do not know if and how $\ce{Cl2O}$ reacts with $\ce{Na2CO3}$. Sorry ! $\endgroup$
    – Maurice
    Feb 15 at 17:19

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