Step 1: Understand the process of dehydration of ethanol.
Dehydration of ethanol involves the removal of a water molecule from ethanol (C\(_2\)H\(_5\)OH) to form an alkene. The reaction is catalyzed by sulfuric acid (\( \text{H}_2\text{SO}_4 \)) and is typically carried out at a temperature of 413 K (around 140°C).
Step 2: Mechanism of the reaction.
At 413 K, ethanol undergoes elimination to form ethene (C\(_2\)H\(_4\)) as the major product. The reaction proceeds through the formation of a carbocation intermediate in the presence of the strong acid \( \text{H}_2\text{SO}_4 \).
The overall reaction is:
\[
\text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{H}_2\text{SO}_4, 413K} \text{C}_2\text{H}_4 + \text{H}_2\text{O}
\]
This forms ethene (ethene is commonly known as ethylene, which is a gas).
Step 3: Identification of the product.
Since the question specifies the reaction conditions, which are typical for ethanol dehydration, the major product is ethyne (C\(_2\)H\(_2\)) after the complete dehydration process, though sometimes, under the specified conditions (with high concentration sulfuric acid and elevated temperature), the final product formed is ethene, which will yield the product \( \text{Ethyne} \).