The reactions are a bit poorly written. While it's true that aqueous H₂CO₃ is produced in this neutralization reaction, the H₂CO₃ rapidly decomposes to yield CO₂(g) and H₂O(l). Writing the product as H₂CO₃(aq) in the net ionic equation is unnecessarily confusing since it portrays the substance as nonionizing yet water-soluble.
In any case, the Na⁺ and the Cl⁻ are the spectator ions here.
I think the correct answer would be the third option. The correct name for the hydrocarbon described above would be 2-heptyne. It has a chemical formula written as CH3 - CH2 - CH2 - CH2 - C ≡ C - CH3. Counting the number of carbons, we have 7 carbon atoms so we use the prefix hepta-. Since it has a triple bond then it is an alkyne. So, it would be named as heptyne. The triple bond is located on the second carbon atom so we write 2 before the name to indicate the location of the triple bond. The name of the compound would be 2-heptyne.
