Answer:
a.
![Fe_2(SO_4)_3(aq)+6NaOH(aq)\rightarrow 2Fe(OH)_3(s)+3Na_2SO_4(aq)](https://img.qammunity.org/2021/formulas/chemistry/college/9o2qmggz04ces2tp6uba15g0zhorogifpn.png)
b.
![2Fe^(2+)(aq)+3(SO_4)^(2-)(aq)+6Na^+(aq)+6OH^-(aq)\rightarrow 2Fe(OH)_3(s)+6Na^+(aq)+3(SO_4)^(2-)(aq)](https://img.qammunity.org/2021/formulas/chemistry/college/wztyx0cpvtabqiqub5nrke23xq1soba6nz.png)
c.
![Fe^(2+)(aq)+3OH^-(aq)\rightarrow Fe(OH)_3(s)](https://img.qammunity.org/2021/formulas/chemistry/college/g2zmiuaun9sis2dfq3ex53jbn0gh6ww8oo.png)
Step-by-step 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.
![Fe_2(SO_4)_3(aq)+6NaOH(aq)\rightarrow 2Fe(OH)_3(s)+3Na_2SO_4(aq)](https://img.qammunity.org/2021/formulas/chemistry/college/9o2qmggz04ces2tp6uba15g0zhorogifpn.png)
Now, for the total ionic equation, we make sure we separate the aqueous species in ions (dissociation) in order to write:
b.
![2Fe^(2+)(aq)+3(SO_4)^(2-)(aq)+6Na^+(aq)+6OH^-(aq)\rightarrow 2Fe(OH)_3(s)+6Na^+(aq)+3(SO_4)^(2-)(aq)](https://img.qammunity.org/2021/formulas/chemistry/college/wztyx0cpvtabqiqub5nrke23xq1soba6nz.png)
Finally, the net ionic equation comes up by cancelling out the spectator ions, those at both reactants and products sides:
![2Fe^(2+)(aq)+6OH^-(aq)\rightarrow 2Fe(OH)_3(s)](https://img.qammunity.org/2021/formulas/chemistry/college/kviegohb3i8wtyw5bppo3z8s8ng5lsi7wv.png)
Or just:
c.
![Fe^(2+)(aq)+3OH^-(aq)\rightarrow Fe(OH)_3(s)](https://img.qammunity.org/2021/formulas/chemistry/college/g2zmiuaun9sis2dfq3ex53jbn0gh6ww8oo.png)
Best regards!