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When $\ce{BaO2}$ reacts with $\ce{HClO},$ the question tells us that that oxygen, barium chloride and one other product is formed.

The equation I came up with is

$$\ce{BaO2 + 2 HClO -> BaCl2 + 2 O2 + 2 H2}\tag{R1}$$

The equation given, however, is

$$\ce{BaO2 + 2 HClO -> BaCl2 + 1 1/2 O2 + H2O}\tag{R2}$$

Why is $\ce{H2O}$ produced and not $\ce{H2}?$ More importantly, what clues/rules are there that would have helped me to predict these products?

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    $\begingroup$ Chloric(I) acid is a strong oxidizer, even stronger than elemental chlorine. Why would it support reduction of $\ce{\overset{+1}{H}^+}$ to $\ce{\overset{0}{H}_2}?$ $\endgroup$ – andselisk Jan 3 at 15:45
  • $\begingroup$ @andselisk Oh right, I see - so this is just a reaction to remember then, rather than to learn any 'rules' from? $\endgroup$ – rdx Jan 3 at 15:56
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    $\begingroup$ Not really, common sense works here all right. One can assume that $\ce{\overset{+1}{Cl}O-}$ is a strong oxidizer even without looking up any reduction potentials simply by noting "unconventional" oxidation number of chlorine. $\endgroup$ – andselisk Jan 3 at 16:00
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    $\begingroup$ TLDR How on Earth would you get a reducing agent from two oxidising ones? $\endgroup$ – Mithoron Jan 4 at 2:03
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    $\begingroup$ Because peroxide is also a reducing agent, forming $\ce{O2}$. Study the half-reactions set up in my answer. $\endgroup$ – Oscar Lanzi Jan 4 at 2:48
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You may have confused barium peroxide with barium metal. The metal can indeed displace hydrogen from hypochlorous acid, not to mention (for a metal as reactive as barium) from the water in which the acid is dissolved; but metal peroxides will produce oxygen and water from the acid. So your missing product is water, not hydrogen.

Tricky half-reactions

If we want to balance this with half-reactions we need to take some care, because we need to simultaneously cancel protons and electrons with a weak acid like hypochlorous acid. Given the reduction half-reaction based on chlorine being reduced from the acid:

$\ce{HClO + H^+ + 2 e^- -> Cl^- + H2O}$ Eq. 1

We must set up the peroxide oxidation to oxygen so that the electron/hydrogen ion ratio is also $2:1$. Thus a proposed reaction

$\ce{BaO2 -> Ba^{2+} + O2 + 2 e ^-}$ Eq. 2

would be balanced but you don't have the proper ratio of electrons to protons matching the weak hypochlorous acid reduction. Add water to get the necessary hydrogen ions, and note that this means you change the oxygen production:

$\ce{BaO2 + H2O -> Ba^{2+} + 2 H^+ + (3/2) O2 + 4 e ^-}$ Eq.3

In other words, water is also oxidized to generate the hydrogen ions necessary to effect the reduction of the weak hypochlorous acid, thus producing that extra half-mole of oxygen per mole of barium peroxide.

Four electrons to two protons in Eq. 3 matches the ratio of two electrons to one proton in Eq. 1. So now we can double Eq. 1 and add to Eq. 3 getting the reaction that avoids both unbalanced protons and unbalanced electrons. As we saw above, one mole of barium peroxide gives 1.5 moles of oxygen in the overall reaction:

$\ce{BaO2 + 2 HClO -> Ba^{2+} + 2Cl^- + (3/2) O2 + H2O}$.

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