When dilute aqueous solution of iron (III) nitrate and sodium hydroxide are mixed, a precipitate is observed. write a molecular, complete ionic, and net ionic equation for the reaction?

##"Fe"_text((aq])^(3+) + 3"OH"_text((aq])^(-) -> "Fe"("OH")_text(3(s]) darr##

You're dealing with a that results in the formation of an insoluble solid that precipitates out of solution.

More specifically, a solution of iron(III) nitrate, ##"Fe"("NO"_3)_3##, will react with a solution of sodium hydroxide, ##"NaOH"##, to form iron(III) hydroxide, ##"Fe"("OH")_3##, which precipitates out of solution.

So, the molecular equation that describes this reaction is

##"Fe"("NO"_3)_text(3(aq]) + 3"NaOH"_text((aq]) -> "Fe"("OH")_text(3(s]) darr + 3"NaNO"_text(3(aq])##

The complete ionic equation will include the solid and all the ions that are present in solution

##"Fe"_text((aq])^(3+) + 3"NO"_text(3(aq])^(-) + 3"Na"_text((aq])^(+) + 3"OH"_text((aq])^(-) -> "Fe"("OH")_text(3(s]) darr + 3"Na"_text((aq])^(+) + 3"NO"_text(3(aq])^(-)##

Notice tha the insoluble solid is not represented as ions!

The net ionic equation will only show the ions that actually take part in the reaction.

The reaction between the two will result in the formation of the solid, which means that the ions that are not a part of the solid will not take part in the reaction.

More specifically, they will act as spectator ions. You can tell which ions are spectator ions by their presence on both sides of the equation.

So, the net ionic equation is

##"Fe"_text((aq])^(3+) + 3"OH"_text((aq])^(-) -> "Fe"("OH")_text(3(s]) darr##

Here's how this reaction would look like

Here the iron(III) nitrate solution is added to the colorless sodium hydroxide solution