Question

How do you balance `Ac(OH)_3(s) -> Ac_2O_3 + 3H_2O`?

Answer 1

It's already partially balanced. All you need is a `2` on `"Ac"("OH")_3(s)`.

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Not much is known about `"Ac"_2"O"_3` or `"Ac"("OH")_3`... but here's what I found.

This reaction is a thermal decomposition that occurs at `1100^@ "C"`. The melting point of `"Ac"_2"O"_3` is estimated to be `1227^@ "C"`.

Thus, we can expect that, if the temperature didn't change significantly from `1100^@ "C"`, the products are solid `"Ac"_2"O"_3` and gaseous `"H"_2"O"`.

Your reaction probably started out like this:

`"Ac"("OH")_3(s) stackrel(Delta" ")(->) "Ac"_2"O"_3(s) + "H"_2"O"(g)`

Given 3 hydrogens on the left, you need to notice that the common multiple of `2` and `3` is `6`. Thus, multiply `3` by `2` and `2` by `3`, meaning that you need two molecules of `"Ac"("OH")_3`.

`2"Ac"("OH")_3(s) stackrel(Delta" ")(->) "Ac"_2"O"_3(s) + "H"_2"O"(g)`

Then, as we just said, you need to multiply `2` by `3`, so you need three water molecules.

`color(blue)(2"Ac"("OH")_3(s) stackrel(Delta" ")(->) "Ac"_2"O"_3(s) + 3"H"_2"O"(g))`

At this point you are done. Tally:

• `"Ac"`: 2 vs. 2
• `"O"`: 2x3 vs. 3+3x1
• `"H"`: 2x3 vs. 3x2

Thus, it is balanced.