Respuesta :
Answer:
a.
[tex]Fe_2(SO_4)_3(aq)+6NaOH(aq)\rightarrow 2Fe(OH)_3(s)+3Na_2SO_4(aq)[/tex]
b.
[tex]2Fe^{2+}(aq)+3(SO_4)^{2-}(aq)+6Na^+(aq)+6OH^-(aq)\rightarrow 2Fe(OH)_3(s)+6Na^+(aq)+3(SO_4)^{2-}(aq)[/tex]
c.
[tex]Fe^{2+}(aq)+3OH^-(aq)\rightarrow Fe(OH)_3(s)[/tex]
Explanation:
Hello!
In this case, since the reaction between sodium hydroxide and iron (III) sulfate yields iron (III) hydroxide, an insoluble base, and sodium sulfate, a soluble salt, we can write the molecular equation as shown below:
a.
[tex]Fe_2(SO_4)_3(aq)+6NaOH(aq)\rightarrow 2Fe(OH)_3(s)+3Na_2SO_4(aq)[/tex]
Now, for the total ionic equation, we make sure we separate the aqueous species in ions (dissociation) in order to write:
b.
[tex]2Fe^{2+}(aq)+3(SO_4)^{2-}(aq)+6Na^+(aq)+6OH^-(aq)\rightarrow 2Fe(OH)_3(s)+6Na^+(aq)+3(SO_4)^{2-}(aq)[/tex]
Finally, the net ionic equation comes up by cancelling out the spectator ions, those at both reactants and products sides:
[tex]2Fe^{2+}(aq)+6OH^-(aq)\rightarrow 2Fe(OH)_3(s)[/tex]
Or just:
c.
[tex]Fe^{2+}(aq)+3OH^-(aq)\rightarrow Fe(OH)_3(s)[/tex]
Best regards!
