One mole of ethyl acetate on treatment with an excess of ${\mathrm{LiAlH}}_4$  in dry ether and subsequent acidification produces

The chemical reaction is as follows.

$\underset{\mathrm{Ethyl} \mathrm{acetate}}{{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5}\underset{\mathrm{Dry} \mathrm{ether}}{\overset{{\mathrm{LiAlH}}_4/{\mathrm{H}}_3\mathrm{O}+}{\to }}{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$

In addition to the alcohol or phenol form from which it was derived, the ester is cleaved to produce a primary alcohol corresponding to the acid portion of the ester.
${\mathrm{LiAlH}}_4$ is used to produce two moles of ethanol from ethyl acetate.
Hence, the correct option (C). 2 moles of ethyl alcohol.