Verifying Reasoning for AQ Solutions and Water's Presence in Net Ionic Equations

[ktpr2]

i would like to see if my reasoning is correct (for aq. solutions):

$$2HBr_(aq_) + 2NH_3_(aq_) -> 2NH_4_(aq_) + Br_2_(aq_)$$
HBR is an acid so it gives H to NH3, which leaves Br all alone. But Br can't exist by itself, so it becoems Br2?
ionic:
$$2H^+_(aq_) + 2Br^-_(aq_) + 2NH^-_3_(aq_) -> 2NH^+_4_(aq_) + 2Br^-_(aq_)$$
net ionic:
$$2H^+_(aq_) + 2NH^-_3_(aq_) -> 2NH^+_4_(aq_)$$

Also, I have a question about water and its presence or lackthereof in net ionic equations (for aq. solutions).

I have
$$2HClO_4_(aq_) + Mg(OH)_2_(s_) -> Mg(ClO_4)_2 (s) + 2H_2O$$
ionic:
(matter states are same as above)$$2H^+ + 2CLO^-_4 + Mg^+^2 + 2OH^- -> Mg(ClO_4)_2 + 2H^+ + 2OH^-$$

but that last part is really $$2H_2O$$. But if i write it like that, then in my net equation i got to keep $$H_2O$$ and the H and OH ... making my net equation the same as my ionic equation. That seems wrong to me. Is there a covention where I can write $$2H_2O$$ and still leave out OH and H as spectator ions in my net ionic equation? Thank you for your time.

 

[GCT]

First check if $Mg(ClO_4)_2$ is a solid, otherwise it would remain hydrated and would not appear in the net ionic equation. And I believe that water will actually be in the net ionic equation, that is $H^+_{(aq)} + OH^-_{(aq)} \xrightarrow{\leftarrow} H_2O_{(l)}$, there's a equilibrium associated with this equation although not on a large scale.

 

[GCT]

that is most of the ionized hydrogen cation and hydroxide will be converted to water.

 

[ktpr2]

ERROR - yeah that mg clo4 is solid, it has (s) in subscript beside it.
EDIT - whoops. That was my assumption. If HCLO is breaking it down, then it'd have to be a liquid, as this kind of reaction goes to completion.

That makes the net equation for (2)
$$2H^+ + 2CLO^-_4 + Mg^+^2 + 2OH^- -> Mg(ClO_4)_2_(aq_) + 2H_2O$$

I take it the reasoning in my first equation involving HBr and NH3 is correct?

 

[GCT]

First, think about it...how is bromine gas even formed? You've got two anions, the two cannot combine unless one of them is a cation. So bromine anion is the final product, usually the ammonium cation interacts ionically with bromine anion to a degree to form a salt adduct (l). However, I don't believe that you need to get this technical here. Thus your net ionic equation seems correct.

 

[ktpr2]

thanks for your help.