# Products of a chemical reaction

1. Oct 22, 2009

### IBY

1. The problem statement, all variables and given/known data

Do anyone know the product of:
$$NaClO+Na_2SO_3$$

The thing is, while searching online, I found:
$$Na_2SO_4+NaCl$$

While the instructor had something along the line of (and he himself isn't sure of the answer, since he is not the one who wrote it):
$$Cl_2+Na_2SO_4+H_2O$$
Or something like that. I don't remember how it went exactly.

The only constant between them is the sodium sulfite. So, which is it?

Last edited: Oct 22, 2009
2. Oct 22, 2009

### Firelion

Hi,
I´m not sure which one of the reactions happens but your second equation cant be right in this form, because you left one of the Na+ out and you still need one Cl-, and one O2-. Where got he the H+ from?

3. Oct 22, 2009

### Staff: Mentor

These are not full reaction equations, these are just products listed.

Cl2 is not the product of hypochlorite reduction, Cl- is. Cl2 is a strong oxidizng agent, so in the presence of SO32- reaction will not stop at Cl2.

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methods

4. Oct 22, 2009

### IBY

So, is this right, then?:
$$NaClO+Na_2SO_3$$--->$$Na_2SO_4+Cl^-$$
It is not balanced, of course, that is something I will do in my own time.

The alternative I got from online research was:
$$NaClO+Na_2SO_3$$--->$$Na_2SO_4+NaCl$$

But I don't know which one is right.

Last edited: Oct 22, 2009
5. Oct 22, 2009

### Staff: Mentor

Your first equation - as written - can't be balanced. Equation is balanced when atoms are balanced and CHARGE is balanced - you can't balance charge having it on one side only.

Also note that what really happens here is a reaction between ClO- and SO32- - Na+ are just spectators.

--
methods

6. Oct 22, 2009

### IBY

Okay, I balanced the redox equation:
$$ClO^-+SO^{-2}_3$$--->$$SO^{-2}_4+Cl^-$$

To:

$$ClO^-+SO^{-2}_3$$--->$$Cl^-+SO^{-2}_4$$

Correction: It looks like after balancing it, things remain the same, since all oxygen is accounted for, there is no need to add water, and so there is no need to balance hydrogen, and charge remains -3 for both sides.

But how does Na come into play?

Last edited: Oct 22, 2009
7. Oct 22, 2009

### Firelion

The Na+ stays as Na+ as long as you dont dry it. So usually (at least )in school you can ignore it in your equation.

8. Oct 22, 2009

### IBY

Thanks!

9. Oct 22, 2009

### Staff: Mentor

That's why it is called a spectator - it just sits there and watches others at work

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