Reduction of ethyl ethanoate, $\mathrm{CH}{}_3-\mathrm{COOC}{}_2\mathrm{H}{}_5$, with excess $\mathrm{LiAlH}{}_4$ in dry ether followed by $\mathrm{H}{}_3\mathrm{O}{}^{+}$ gives

$\mathrm{LiAlH}{}_4$ reduces an ester fully to two alcohols: the acyl part becomes a primary alcohol and the alkoxy part is released as alcohol. Here the acyl part $\mathrm{CH}{}_3\mathrm{CO}$ gives $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{OH}$ and the $\mathrm{OC}{}_2\mathrm{H}{}_5$ part is also ethanol. Hence only ethanol is formed.