Question

balance the following chemical equation:
fe(oh)₃ → fe₂o₃ +
h₂o

Explanation:

Step1: Balance Fe atoms

On the left, we have 1 Fe in $\ce{Fe(OH)3}$. On the right, we have 2 Fe in $\ce{Fe2O3}$. So we put a coefficient of 2 in front of $\ce{Fe(OH)3}$:
$2\ce{Fe(OH)3}
ightarrow \ce{Fe2O3} + \square\ce{H2O}$

Step2: Balance O and H atoms

Now, left side has $2\times3 = 6$ O from $\ce{OH^-}$ and 2 Fe. Wait, let's count H and O. After step 1, left has $2\times3 = 6$ H (from $\ce{OH}$) and $2\times3 = 6$ O (from $\ce{OH}$) plus 0 from Fe? Wait, $\ce{Fe(OH)3}$ has 3 O per formula, 2 formulas: 6 O, 6 H. Right side: $\ce{Fe2O3}$ has 3 O, and $\ce{H2O}$ has 1 O per molecule. Let's count H: left has 6 H (from $2\ce{Fe(OH)3}$), so right $\ce{H2O}$ must have 6 H, so coefficient for $\ce{H2O}$ is 3 (since $3\times2 = 6$ H). Now check O: left has $2\times3 = 6$ O (from $\ce{OH}$). Right: $\ce{Fe2O3}$ has 3 O, $\ce{H2O}$ has $3\times1 = 3$ O. Total 6 O. Perfect. So the equation is $2\ce{Fe(OH)3}
ightarrow \ce{Fe2O3} + 3\ce{H2O}$

Answer:

The coefficients are 2 (for $\ce{Fe(OH)3}$) and 3 (for $\ce{H2O}$), so the balanced equation is $\boldsymbol{2\ce{Fe(OH)3}
ightarrow \ce{Fe2O3} + 3\ce{H2O}}$