How much of Sodium dichloro-s-triazinetrione dihydrate should I use to replace bleach? [closed]








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I am trying to determine how much sodium dichloro-s-triazinetrione dihydrate powder I should add to water to create a dilution that can be injected into my water supply to react with the Hydrogen Sulfide and remove it.


I have a system in my home that uses sodium hypochlorite (bleach) and water. There is a 30 gallon water barrel that we mix 1 gallon of bleach (5.25% Sodium hypochlorite) with 29 gallons of water. I believe that means the resulting mixture is 0.16666% Sodium hypochlorite. That is injected using a small pump into the water supply where it mixes in a tank and oxidizes the Hydrogen Sulfide.


I want to stop buying bottles of bleach which is expensive and contains additives other than sodium hypochlorite. I would like to switch to sodium dichloro-s-triazinetrione dihydrate which I already have an abundance of in the form of Spa Sanitizer. The Spa Sanitizer label indicates it is 99.99% pure sodium dichloro-s-triazinetrione dihydrate. 


So how much of the sodium dichloro-s-triazinetrione dihydrate powder do I need to mix with 29 gallons of water to create a solution similar to the one created by the bleach? (0.16666% sodium hypochlorite)


Or, maybe what I should be asking is what is the amount of free chlorine in the 0.16666% sodium hypochlorite solution and how much sodium dichloro-s-triazinetrione dihydrate powder do I need to mix with 29 gallons of water to achieve the same amount of free chlorine? Since I believe the free chlorine in sodium hypochlorite and sodium dichloro-s-triazinetrione dihydrate are not actually the same.




According to a relevant reference (Ref.1) Following reactions happen when sodium hypochlorite and sodium dichloro-s-triazinetrione dissolve in water, respectively:


$$\ce{NaOCl + H2O -> HOCl + NaOH} \tag {1}$$


$$\ce{NaCl2(NCO)3 + 2 H2O <=> 2 HOCl + NaH2(NCO)3} \tag {1}$$


Hence it is safe to assume that you need only $\pu{0.5 mol}$ of $\ce{NaCl2(NCO)3}$ (sodium dichloro-s-triazinetrione) or in your case $\ce{NaCl2(NCO)3.2H2O}$ (sodium dichloro-s-triazinetrione dihydrate), in place of each $\pu{1.0 mol}$ of $\ce{NaOCl}$ used. Let's see how much of $\ce{NaOCl}$ in $\pu{1.0 US gal}$ of $5.25\%$ bleach solution:


$$5.25\%(w/v) = \\ 
 \frac{\pu{5.25 g} \;\ce{NaOCl}}{\pu{100 mL} \;\text{solution}}\times \frac{\pu{1.0 mol} \;\ce{NaOCl}}{\pu{74.44 g} \;\ce{NaOCl}}\times \frac{\pu{1000 mL} \;\text{solution}}{\pu{1.0 L} \;\text{solution}}\times \frac{\pu{3.79 L} \;\text{solution}}{\pu{1.00 US gal} \;\text{solution}}\\
 = \pu{2.673 mol/US gal}$$


As states above, you need only half of this amounts in solid $\ce{NaCl2(NCO)3.2H2O}$. You can calculate the amount of $\ce{NaCl2(NCO)3.2H2O}$ in grams as follows:


$$\pu{1 US gal} \: \text{of } 5.25\%(w/v) = \\
\pu{2.673 mol} \; \text{of } \ce{NaOCl} \times \frac{\pu{0.5 mol} \; \ce{NaCl2(NCO)3.2H2O}}{\pu{1.0 mol} \; \ce{NaOCl}} \times \frac{\pu{255.98 g} \;\ce{NaCl2(NCO)3.2H2O}}{\pu{1.0 mol} \;\ce{NaCl2(NCO)3.2H2O}} \\
= \pu{342.1 g} \;\ce{NaCl2(NCO)3.2H2O}$$


Thus, you can dissolve $\pu{342.1 g}$ of $\ce{NaCl2(NCO)3.2H2O}$ in your 30-gal (US) water barrel to get the same effect of 1-US gal of 5.25% bleach solution. Keep in mind that rate of chlorine release from $\ce{NaCl2(NCO)3.2H2O}$ is slower than that of bleach solution (Ref.1).




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**Late Edit:**


This is about comment made by OP on my post. In that comment, OP states that the cost for dichloro-s-triazinetrione dihydrate (Di-Chlor) and beach are almost same. But I do not agree with that. Just for curiosity, I checked the prices for two chemicals: You can buy $\pu{50 lbs}$ bucket of Di-Chlor granules [online] from [Pool Supply World](https://www.poolsupplyworld.com/PoolSupplyWorld-Di-Chlor-Granular-Stabilized-Chlorine-Buckets/granular-chlorine-buckets.htm?Specialty=Stabilizers&Container%20Size=Bucket%A050%20lb&Active%20Ingredient=Dichlor-S-Trianzinetrione&Chemical%20Size=Granular&Product%20Type=Pool%20Chemical&Chemical%20Type=Chlorine&Manufacturer=PoolSupplyWorld&Online%20Favorite=2018&utm_campaign=smart%5Fshopping%5Fchemicals&ecmpid=aw%5Fpla%5Fsearch%5FPSW%2DCSC168&gclid=EAIaIQobChMIx7PO6eir5wIVC9vACh2Vagp9EAQYASABEgLUUfD%5FBwE&utm_source=google&utm_medium=cpc&xzm2=&adtype=paid) for $\pu{132.99 USD}$; and You can buy $\pu{121 fl \: OZ}$ container of 6% bleach solution from [Target](https://www.target.com/p/clorox-disinfecting-bleach-regular-121oz/-/A-14711215?ref=tgt_adv_XS000000&AFID=google&fndsrc=tmnv&DFA=71700000049859076&CPNG=PLA_DVM%2B0060H00000n8n7nQAA-Clorox-Laundry-Monthly+Search-2019-Flight-AO&adgroup=PLA_Clorox-Laundry&LID=700000001393753pgs&network=g&device=c&location=1022639&gclid=EAIaIQobChMIws3V8Oqr5wIVSb7ACh00-wXgEAQYASABEgJJP_D_BwE&gclsrc=aw.ds) for $\pu{4.49 USD}$ (let's assume its about $\pu{1 US\:Gal}$ of 5.25% bleach solution). According to my calculations above, $\pu{50 lbs}$ bucket of Di-Chlor granules equal to:


$$\pu{50.0 lbs} \; \ce{NaCl2(NCO)3.2H2O} \times \frac {\pu{454 g}}{\pu{1.0 lbs}} \times \frac{\pu{1.0 US\:Gal} \;\text{5.25% }\ce{NaClO}}{\pu{342.1 g} \;\ce{NaCl2(NCO)3.2H2O}}\\
 \approx \pu{66.5 US\:Gal} \text{ of 5.25% }\ce{NaClO} \text{ solution} $$


As a consequence, the cost for $\pu{50 lbs}$ bucket of Di-Chlor granules is $\pu{132.99 USD}$ while the cost for $\pu{66 US\:Gal}$ of 5.25% bleach solution containers is $\pu{4.49 USD} \times 66 \approx \pu{296.34 USD}$ (more than twice the cost for Di-Chlor granules needed).




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**References:**


1. Thomas Clasen, Paul Edmondson, "Sodium dichloroisocyanurate (NaDCC) tablets as an alternative to sodium hypochlorite for the routine treatment of drinking water at the household level," *International Journal of Hygiene and Environmental Health* **2006**, *209(2)*, 173-181 (<https://doi.org/10.1016/j.ijheh.2005.11.004>).


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$1$ mole of sodium dichloro-s-triazinetrione weighs $220$ g, and is able to liberate $2$ moles of $Cl\_2$ in water. If you use the same substance as dehydrate, one mole will weigh $256$ g.
By comparison, $1$ mole of bleach $NaClO$ ($74.5$ g) is able to yield $1$ mole $Cl\_2$ in water. 


So if you know the mass *m* of $NaClO$ that you are adding to your water, you may obtain the same effect in solution by using a mass *m'* of sodium dichloro-s-triazinetrione dihydrate equal to *m'* = (*m*/$74.5$)·($256/2$) g = $1.71$·*m*. 


This means that you must use much more dichloro-s-triazinetrione than bleach.


In an aside I wonder why you would renounce to using bleach. It is cheap and efficient. Apart from $NaClO$ the main "additive" present in bleach is $NaCl$, or cooking salt, which is not a real nuisance.


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