Question #4d141
1 Answer
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,
#"Fe"("NO"_ 3)_ (3(aq)) -> "Fe"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-)#
#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#
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
#"Fe"("NO"_ 3)_ (2(aq)) + color(red)(3)"NaOH"_ ((aq)) -> "Fe"("OH")_ (3(s)) darr + 3"NaNO"_ (3(aq))#
The complete ionic equation looks like this
#"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"_ ((aq))^(-)#
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
#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
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#