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

Apr 12, 2016

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

Not much is known about ${\text{Ac"_2"O}}_{3}$ or "Ac"("OH")_3... but here's what I found.

This reaction is a thermal decomposition that occurs at ${1100}^{\circ} \text{C}$. The melting point of ${\text{Ac"_2"O}}_{3}$ is estimated to be ${1227}^{\circ} \text{C}$.

Thus, we can expect that, if the temperature didn't change significantly from ${1100}^{\circ} \text{C}$, the products are solid ${\text{Ac"_2"O}}_{3}$ and gaseous $\text{H"_2"O}$.

Your reaction probably started out like this:

$\text{Ac"("OH")_3(s) stackrel(Delta" ")(->) "Ac"_2"O"_3(s) + "H"_2"O} \left(g\right)$

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.

$2 \text{Ac"("OH")_3(s) stackrel(Delta" ")(->) "Ac"_2"O"_3(s) + "H"_2"O} \left(g\right)$

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

$\textcolor{b l u e}{2 \text{Ac"("OH")_3(s) stackrel(Delta" ")(->) "Ac"_2"O"_3(s) + 3"H"_2"O} \left(g\right)}$

At this point you are done. Tally:

• $\text{Ac}$: 2 vs. 2
• $\text{O}$: 2x3 vs. 3+3x1
• $\text{H}$: 2x3 vs. 3x2

Thus, it is balanced.