[ldixon]

Ok here is the problem:
2NaI (aq)  +  Cl2 (g) -----   2NaCl (aq)  +  I2 (s)

For the net ionic I got:
2I- (aq)  +  Cl2 (g)  -------  2Cl- (aq)  +  I2 (s)


Is this correct?  Most chlorides and iodides are soluble, but one states (g) and the other states (s) so does that mean they are not soluble? 

[english]

Net ionic removes all spectators.  Anything soluble is a spectator.  A gas is being formed, so it is being released from solution, thus is not soluble.

Be careful, some of your species you have in your equation are soluble and shouldn't be there.

[ldixon]

2Na+(aq)  +  2I-(aq)  +  2Cl-(aq)  >  2Na+(aq)  +  2Cl-(aq)  +  2I-(s)
 
Na+, Cl-
 
I-(aq)  >  I-(s)
 
Would it be more like this?  Im not sure what you mean about the Cl (g)?

[english]

Oh sorry I screwed up.  I didn't pay attention to what was going on!

Ignore what I said earlier.

[ldixon]

So the Cl2 gas and the 2Cl- (aq) would cancel each other out?  I thought all net ionic equations delt with soluble/insulble?   

[english]

So the Cl2 gas and the 2Cl- (aq) would cancel each other out? I thought all net ionic equations delt with soluble/insulble?

No. 

No cancelling.

The Cl2 gas must be soluble because there is no other way it could become solvated as a product.

Our reaction is just the precipitation of I2:

2I ^-_(aq) ----> I_2(s)

[english]

I thought all net ionic equations delt with soluble/insulble?   

Yes.  Don't stop thinking this way I just screwed up.

Human is all 

[ldixon]

2Na+(aq)  +  2I-(aq)  +  2Cl-(aq)  >  2Na+(aq)  +  2Cl-(aq)  +  2I-(s)  complete ionic equation

Na+, Cl- spectator ions

2I-(aq)  >  I2(s) net ionic equation

So if I put this answer on a test it would be correct?


Thanks so much for the *delete me*

[english]

You have to designate (g) for Cl2.  It becomes solvated but is a gas before the reaction occurs.

[ldixon]

Can anybody give me a second opinion on this?  I have received mixed answers, but I dont have a solid answer.

[AWK]

2I-(aq)  >  I2(s) net ionic equation


Thanks so much for the *delete me*

This  is only  a half-reaction. Full ionic reaction is below:

2I^-(aq) + Cl₂(g)   > Cl^-(aq) +  I2(s)
and you was correct from the same beginning

[english]

2I-(aq)  >  I2(s) net ionic equation


Thanks so much for the *delete me*

This  is only  a half-reaction. Full ionic reaction is below:

2I^-(aq) + Cl₂(g)   > Cl^-(aq) +  I2(s)
and you was correct from the same beginning

Thanks AWK I knew I screwed up somehwhere.