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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.

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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).


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 for $\pu{132.99 USD}$; and You can buy $\pu{121 fl \: OZ}$ container of 6% bleach solution from Target 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).


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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    $\begingroup$ Your answer makes sense. I also found this link oxy.com/OurBusinesses/Chemicals/Products/Documents/… since asking and on page 4 it has conversion chart showing that 1lb of NaCl2(NCO)3⋅2H2O is approximately equal to 1.2 gal of 5.25% bleach. Your equations roughly support this chart and vise versa. I appreciate knowing the why behind the chart and you gave me that. It is in the end worth noting that based on the cost of the shock and the bleach they are about equal for this use. But i have the spa shock already. Thanks. $\endgroup$ – Karl Bystrak Jan 30 at 0:10
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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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