On the molybdenum(III) oxide-catalysed oxidation of alkanes to carbonyls The following reaction is specified in my notes;
$$\ce{CH4 -&gt;[Mo2O3, \Delta]HCHO}$$ The intermediate steps involve the dehydration of an alcohol to yield the final product.
Wouldn't this only work if there's acid in the reaction mixture? Or is it that water functions as a proton donor, inducing the protonation of the hydroxyl and causing the dehydration?
The above depicts a concept, rather than an equation for that there is no balance of atoms on the left and right of the arrow.  Because if $\ce{Mo2O3}$ is said to be used as catalyst rather than a (stoichiometric) reagent, then the oxygen in formaldehyde must be of different origin than $\ce{Mo2O3}$.
On industrial scale, formaldehyde is prepared from methanol by an  heterogeneous redox-reaction.  The oxygen needed may come from oxygen gas, or simply air.  Thus, the last step known as Formox process is
$$ \ce{2 CH3OH + O2 -&gt; 2 CH2O + 2 H2O} $$
(from wikipedia)
$\ce{Mo2O3}$ is deployed ease the advancement of the reaction, i.e., to perform the wanted reaction at lower temperature (heating costs money) than without this help, and to further increase rate or / and yield of said reaction compared performing this conversion at room temperature.