Answer:
[tex]HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)[/tex]
Explanation:
Hello there!
In this case, for this neutralization reaction, it is possible to realize that one the neutralization products is water (pH=7) and the other one is the salt coming up from the cation of the NaOH and the anion of the HI:
[tex]HI(aq)+NaOH(aq)\rightarrow NaI+H_2O[/tex]
Moreover, since the solubility of NaI is large in water, we infer it remains aqueous whereas the water is maintained as liquid:
[tex]HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)[/tex]
Which is also balanced as the number of atoms of all the elements is the same at both sides.
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