Question

What type of chemical reaction does the reaction belong to? A piece of sodium metal is added to absolute ethanol to form sodium ethoxide and hydrogen gas.
A) Precipitation Reactions
B) Neutralization Reactions
C) Decomposition Reactions
D) Displacement Reaction

Answer 1

Sodium is metal and ethanol is alcohol, both react to form sodium ethoxide and hydrogen gas is produced. If we look at the structures of both the compounds, reactant, and product, we can observe that some atom is being removed from its place and substituted by another one.

Complete answer:
The chemical reaction for the reaction of sodium metal and ethanol to form sodium ethoxide and hydrogen gas will be - $2Na + 2{C_2}{H_5}OH \to 2{C_2}{H_5}ONa + {H_2}$
Now as the reaction explains, the hydrogen atom from ethanol is removed and sodium is substituted in its place. According to the reaction, two moles of sodium will react to two moles of ethanol to produce two moles of sodium ethoxide and one mole of hydrogen gas.
This is surely not a precipitation reaction as there is no precipitate formation. Salts are usually the end product of the neutralization reaction, as there is no salt in the product, the given reaction is not a neutralization reaction. In a decomposition reaction, the reactant is broken into smaller compounds which is not the case here. So the given reaction is not an example of a decomposition reaction.
As there is the displacement of the hydrogen atom by the sodium atom, so the given reaction is an example of a displacement reaction.
Hence, the correct answer is D) displacement reaction.

Note:
Another way of writing the given reaction is - $Na + {C_2}{H_5}OH \to {C_2}{H_5}ONa + \dfrac{1}{2}{H_2}$ . Here the stoichiometric coefficients are changed. In the previous reaction, divide all the stoichiometric coefficients by two and you will get the reaction - $Na + {C_2}{H_5}OH \to {C_2}{H_5}ONa + \dfrac{1}{2}{H_2}$