Moles of Reduction Product in the Oxidation of Aluminium by Nitric Acid
You simply build the half reactions and add them up so you can see how many moles of hydrogen gas are produced by a mole of Aluminium being oxidised. (Note: Your first equation is unbalanced.)
\begin{aligned}\ce{
(1)&amp;&amp; Al + 6H2O &amp;-&gt; Al(OH2)_{6}^{3+} + 3e^{-}&amp;|\cdot2\\
+ (2)&amp;&amp; 2H3O+ + 2e^{-} &amp;-&gt; H2 ^ + 2H2O &amp;|\cdot3\\\hline
(3)&amp;&amp; 2Al + 12H2O + 6H3O+ &amp;-&gt; 2Al(OH2)_{6}^{3+} + 3H2 ^ + 6H2O\\
(4)\equiv(3)&amp;&amp; 2Al + 6H2O + 6H3O+ &amp;-&gt; 2Al(OH2)_{6}^{3+} + 3H2 ^ \\
(5)\equiv(3)&amp;&amp; Al + 3H2O + 3H3O+ &amp;-&gt; Al(OH2)_{6}^{3+} + \color{\green}{3/2}H2 ^ \\
}\end{aligned}

There are $\color{\green}{\text{one and a half}}$ mole Hydrogen gas formed for each mole of Aluminium.
Hence, three mole of Hydrogen gas are formed by two mole of Aluminium. 



Edit upon Dissenters follow up question "Why exactly would hydronium ion be the source?"
It says in the question that the solution is $2\mathrm{M}~\ce{HONO2}$ (nitric acid, although this is a strange way of writing it). Therefore there will be an excess of $\ce{H3O+}$. And cations travel to the cathode. Water could be a proton source, but not in that scenario.

It is necessary to acidify the solution, because otherwise you would passify the aluminium surface with $\ce{Al(OH)3}$