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According to Wikipedia, molar mass of anhydrous sodium dichloroisocyanurate is approximately 220 g per mole, and molar mass of Cl2 is 71 g per mole. Therefore there is 323 mg of chlorine in 1 g of NaDCC. But according to this page, 33 mg of NaDCC releases 20 mg of chlorine, and so there is 606 mg of chlorine in 1 g of NaDCC. Where have I made a mistake?

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    $\begingroup$ You're wrongfully assuming the amount of chlorine released is the same as the amount of chlorine contained in NADCC, but it does not dissociate completely. $\endgroup$
    – Sam202
    Oct 29, 2022 at 22:55
  • $\begingroup$ By my calculations 33 mg of NaDCC would release 10.65 mg of chlorine, which is almost two times less than what that page claims. $\endgroup$
    – Karabin
    Oct 29, 2022 at 23:06
  • $\begingroup$ In your page, "chlorine" isn't referring to $\ce{Cl2}$ but rather to $\ce{HOCl}$, which is released when added to water. $\endgroup$
    – Sam202
    Oct 29, 2022 at 23:39
  • $\begingroup$ Assuming that all of the chlorine goes from NaDCC to HOCl, 33 mg of NaDCC would result in 15.75 mg of HOCl, which is still less than what is claimed on that page. $\endgroup$
    – Karabin
    Oct 30, 2022 at 0:49
  • $\begingroup$ $\ce{HOCl}$ also partially dissociates into $\ce{OCl^-}$ ions, which are also accounted for in the 20 mg of chlorine produced. $\endgroup$
    – Sam202
    Oct 30, 2022 at 3:21

1 Answer 1

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I believe the discrepancy comes with the $\ce{Na+}$. Each NaDCC has that positive ion. Since there is one $\ce{Na+}$ and two $\ce{Cl}$ per NaDCC. What I suspect happens is that there is that the page derives its values from calculating the abundance of two species: $\ce{NaClO}$ rather than $\ce{ClO-}$ and $\ce{HClO}$. The exact ratio between the two may not be 1:1, but I believe that the mass of all the $\ce{Na+}$ was included in the calculation of the amount of active chlorine.

Revising the calculation, you'd get about $\pu{19.0g}$ of chlorine species including $\ce{NaOCl}$ and $\ce{HOCl}$ with the following expression below (assuming the two species are in 1:1 ratio):

$$\frac{\pu{33 mg}}{\pu{220 g/mol}}\cdot MM(\ce{NaOCl)} + \frac{\pu{33 mg}}{\pu{220 g/mol}}\cdot MM(\ce{HOCl)} = \pu{19.0 g/mol}$$

This is much closer to what was reported. Note that the ratio of the two species is likely not 1:1. However, this would give the largest possible value as all the $\ce{Na+}$ is used up from NaDCC. I am not too sure how got and marketed the value of $\pu{20g}$, but I am guessing it has to do with what I mentioned above and maybe some slight bias in rounding.

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