# Question 4d141

Jun 7, 2017

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

#### Explanation:

For starters, it's worth mentioning that you're dealing with ionic compounds, so those are not molecules, they are formula units.

That said, you don't need to know how many formula units of each reactant are mixed together to write the balanced chemical equation.

Iron(III) nitrate and sodium hydroxide are soluble in water, which implies that they exist as ions in aqueous solution,

${\text{Fe"("NO"_ 3)_ (3(aq)) -> "Fe"_ ((aq))^(3+) + 3"NO}}_{3 \left(a q\right)}^{-}$

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

When these two solutions are mixed, the iron(III) cations will combine with the hydroxide anions to form iron(III) hydroxide, an insoluble solid that precipitates out of solution.

The second product of the reaction is aqueous sodium nitrate.

The balanced chemical equation looks like this

${\text{Fe"("NO"_ 3)_ (2(aq)) + color(red)(3)"NaOH"_ ((aq)) -> "Fe"("OH")_ (3(s)) darr + 3"NaNO}}_{3 \left(a q\right)}$

The complete ionic equation looks like this

${\text{Fe"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-) + color(red)(3)"Na"_ ((aq))^(+) + color(red)(3)"OH"_ ((aq))^(-) -> "Fe"("OH")_ (3(s)) darr + 3"Na"_ ((aq))^(+) + 3"OH}}_{\left(a q\right)}^{-}$

The net ionic equation, which you get by eliminating the spectator ions, i.e. the ions that are present on both sides of the equation, looks like this

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

Now, my guess is that this is supposed to be a limiting reagent problem.

If that is the case, you must use the $1 : \textcolor{red}{3}$ mole ratio that exists between the two reactants to find the limiting reagent. Pick one of the two reactants and calculate the number of formula units needed to ensure that all the formula units of this first reactant take part in the reaction.

30 color(red)(cancel(color(black)("f. units Fa"("NO"_3)_3))) * (color(red)(3)color(white)(.)"f units NaOH")/(1color(red)(cancel(color(black)("f. unit Fa"("NO"_3)_3)))) = "90 f units NaOH"

This tells you that in order for all the molecules of iron(III) nitrate to react, you need $40$ formula units of sodium hydroxide. Sine you only have $15$ formula units of this reactant, you can say that sodium hydroxide will act as a limiting reagent.

Sodium hydroxide will be completely consumed and the reaction will produce

15 color(red)(cancel(color(black)("f. units NaOH"))) * ("1 f. unit Fe"("OH")_3)/(color(red)(3)color(red)(cancel(color(black)("f. units NaOH")))) = "5 f. units Fe"("OH")_3#