[Chris1]

In inorganic chemistry we had a lab work in which we synthesized barium iodate. One of the complementary tests was to heat it. We observed that iodine was formed, but I can't decide which equation is the correct one for this reaction, I have two options:
5Ba(IO₃)₂ → 9O₂ + 4I₂ + Ba₅(IO₆)₂
2Ba(IO₃)₂ → 2BaO + 2I₂ + 5O₂
I remember that in class we discussed that IO₃^- disproportionates, but on the other hand in the second reaction IO₃^- is reduced and the oxygen is oxidized, so I don't see why this reaction is impossible.
Which of these two reactions actually took place? And why?

[...]

The first one, because you will get a double salt BaI₂·4BaO₃ by heating.
Keyword: Rammelsberg reaction.

[AWK]

Ba₂(IO₆)₂ - barium orthoperiodate has nothing common with a double salt.

[...]

Ba₂(IO₆)₂ [...]

Ba₅(IO₆)₂

[...] barium orthoperiodate has nothing common with a double salt.

Correct. But what I'm trying to say is that the product of this reaction should be a double salt, because this is the definition of the Rammelsberg reaction...

[AWK]

Thank you for correction my printing error.

Such a formula could have been proposed by Carl Friedrich Rammelsberg in 1838 - other times, a different notation and understanding of chemistry. But it is not right now to give this ancient information as true if we know the true course of this reaction and we also know the structure of the product of this reaction.

Rammelsberg is trying to match the Ba⁵J²O¹² formula for various oxygen combinations, e.g. Ba⁵O⁵ + J²O⁷ or BaJ² + Ba⁴O¹² (original form of writing at that time)

Moreover, both reactions given by Chris1 are true: the former at ~580°C to 800°C, the later above 800 C