# A solution of iron(III)chloride is mixed with a solution of sodium hydroxide and reacts to yield solid iron (III) hydroxide and aqueous sodium chloride. What is the balanced chemical equation for this reaction?

Jun 14, 2017

${\text{FeCl"_ (3(aq)) + 3"NaOH"_ ((aq)) -> "Fe"("OH")_ (3(s)) darr + 3"NaCl}}_{\left(a q\right)}$

#### Explanation:

Start by writing the unbalanced chemical equation that describes this double displacement reaction

$\text{FeCl"_ (3(aq)) + "NaOH"_ ((aq)) -> "Fe"("OH")_ (3(s)) darr + "NaCl}$

In order to balance this chemical equation, you can use the fact that the two reactants and sodium chloride are soluble in water, which implies that they exist as ions in aqueous solution.

${\text{FeCl"_ (3(aq)) -> "Fe"_ ((aq))^(3+) + 3"Cl}}_{\left(a q\right)}^{-}$

${\text{NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

${\text{NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

You can rewrite the unbalanced chemical equation as

${\text{Fe"_ ((aq))^(3+) + 3"Cl"_ ((aq))^(-) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "Fe"("OH")_ (3(s)) darr + "Na"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

The first thing that stands out here is that you need to balance the chloride anions by multiplying the chloride anions present on the products' side by $\textcolor{b l u e}{3}$.

Keep in mind that the chloride anions are part of the sodium chloride, so if you multiply the chloride anions by $\textcolor{b l u e}{3}$, you must do the same for the sodium cations

${\text{Fe"_ ((aq))^(3+) + 3"Cl"_ ((aq))^(-) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "Fe"("OH")_ (3(s)) darr + color(blue)(3)"Na"_ ((aq))^(+) + color(blue)(3)"Cl}}_{\left(a q\right)}^{-}$

Now balance the sodium cations present on the reactants' side by multiplying them by $\textcolor{red}{3}$. Once again, the sodium cations are part of the sodium hydroxide, so make sure to multiply the hydroxide anions as well!

${\text{Fe"_ ((aq))^(3+) + 3"Cl"_ ((aq))^(-) + color(red)(3)"Na"_ ((aq))^(+) + color(red)(3)"OH"_ ((aq))^(-) -> "Fe"("OH")_ (3(s)) darr + color(blue)(3)"Na"_ ((aq))^(+) + color(blue)(3)"Cl}}_{\left(a q\right)}^{-}$

The hydroxide anions are now balanced because you have $3$ hydroxide anions on the products' side as part of the insoluble iron(III) hydroxide. The same can be said about the iron(III) cations, so put all this together

$\left[{\text{Fe"_ ((aq))^(3+) + 3"Cl"_ ((aq))^(-)] + color(red)(3) xx ["Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)] -> "Fe"("OH")_ (3(s)) darr + color(blue)(3)xx["Na"_ ((aq))^(+) + "Cl}}_{\left(a q\right)}^{-}\right]$

to get the balanced chemical equation

${\text{FeCl"_ (3(aq)) + color(red)(3)"NaOH"_ ((aq)) -> "Fe"("OH")_ (3(s)) darr + color(blue)(3)"NaCl}}_{\left(a q\right)}$

You could also write the net ionic equation, which does not include the spectator ions, i.e. the ions that are present on both sides of the balanced chemical equation.

"Fe"_ ((aq))^(3+) + color(red)(3)"OH"_ ((aq))^(-) -> "Fe"("OH")_ (3(s)) darr