Question

4. Sodium sulfite reacts with iron(III) nitrate to produce sodium sulfate and iron(II) nitrate.

Write a balanced net ionic equation for this reaction when it takes place in the
presence of excess OH- ions.

Answer 1

okay so do yu know your solubitity rules?

Answer 2

no

Answer 3

http://www.csudh.edu/oliver/chemdata/solrules.htm

Answer 4

Sodium sulfite reacts with iron(III) nitrate to produce sodium sulfate and iron(II) nitrate.

\[Na _{2}SO _{3} + FeNO _{3} -> Na _{2} SO _{4} + FeNO _{2}\]

Answer 5

Okay.

Answer 6

From what you have said you get the following:
Na2SO3 + Fe(NO3)3 --> Na2(SO4)+Fe(NO3)2
However you we have to balance it and we get the following:
3Na2SO3 + 2Fe(NO3)3 --> 3Na2(SO4)+3Fe(NO3)2
Now assuming that the reactants are in its ionic form and the products are in its ionice form and are dissolved in the solution: we get the following equation:
6Na(+) + 3SO3(-2) + 2Fe(3+) +6NO3(-) --> 6Na(+) + 3SO4(-2) + 3Fe(2+) + 6NO3(-)
However we need the net ionic equation, this implies that we have to remove those that are not taking part in the reaction, they are known as spectators.
6Na(+) and 6NO3(-) doesn't take affect the outcome of the reaction so we can remove them from the equation and get the net ionic equation:

This is the net ionic equation:
3SO3(-2) + 2Fe(3+) --> 3SO4(-2)+3Fe(2+)

I hope it was helpful brother, I don't know if I have made any mistakes regarding balancing the equation, but If I have please correct.

Answer 7

Thank you! Both of you!!