What is the main product of the reduction of ethyl ethanoate with lithium aluminum hydride?

The reactant is ethyl ethanoate, which is an ester with the formula $\text{CH}_3\text{COOCH}_2\text{CH}_3$.
The reagent is lithium aluminum hydride ($\text{LiAlH}_4$), a strong reducing agent.

$\text{LiAlH}_4$ reduces esters to primary alcohols. The ester linkage ($\text{-COO-}$) is cleaved, and both the acyl part and the alkoxy part are reduced to alcohols.

The acyl part ($\text{CH}_3\text{CO}-$) of ethyl ethanoate is reduced to ethanol ($\text{CH}_3\text{CH}_2\text{OH}$).
The alkoxy part ($-\text{OCH}_2\text{CH}_3$) of ethyl ethanoate is also reduced to ethanol ($\text{CH}_3\text{CH}_2\text{OH}$).

The overall reaction is:

Therefore, the product obtained is ethanol.

The final answer is $\text{A. ethanol}$.