Why do you need excess LAH to reduce esters?
Lithium aluminium hydride or LAH is the more everyday form of $\ce{LiAlH4}$. When originally produced, most of it is indeed $\ce{LiAlH4}$ but that quickly partially hydrolyses giving the grey powder you may or may not know from the lab. As such, $\ce{LiAlH4}$ is practically never entered into reactions in pure form. Since it is impure, excess reagent is added to make sure the reduction job is performed as desired.

Also, remember to consider the reaction meachnism:

$$\ce{R-COOR' -&gt;[LAH][- HOR'] R-CHO -&gt;[LAH] R-CH2OH}\tag{1}$$

Note that the second step, the reduction of an aldehyde, is more rapid than the first as aldehydes are more reactive towards nucleophiles than esters. Thus, if you attempted to bargain, hoping to arrive at the aldehyde selectively, you would end up with under half of your reactant reduced to the alcohol and half of it still as an ester. Another reason to always add excess just to be sure.