Neutralisation Reaction Confusion
First, let's assume these reaction happen in aqueous solution and everything is soluble. The two reaction you are interested in are:

$$\ce{H2SO4(aq) + NaOH(aq) -&gt; HSO4-(aq) + H2O(l) + Na+(aq)}\label{rxn:1}\tag{1}$$

$$\ce{H2SO4(aq) + 2NaOH(aq) -&gt; SO4^2-(aq) + 2H2O(l) + 2Na+(aq)}\label{rxn:2}\tag{2}$$

Your saying that reaction \eqref{rxn:2} might happen even when $\ce{NaOH}$ is limiting, and you would end up with some sulfuric acid and some sulfate. In a thought experiment, let's say that is what happened. Now, we can discuss whether sulfuric acid and sulfate, which are present in equimolar amounts, will react with each other according to:

$$\ce{H2SO4(aq) + SO4^2-(aq) &lt;=&gt; 2HSO4-(aq)}\label{rxn:3}\tag{3}$$

Sulfuric acid is a strong acid, and sulfate is a weak (very weak) base. This is like mixing hydronium ($\ce{H3O+}$) and hydroxide ($\ce{OH-}$) ions, which (mostly) form water. So just like water, which occurs mostly undissociated, the equilibrium of reaction \eqref{rxn:3} lies on the side of the products when the reactants are present in equimolar amounts. 

Of course, if you add more $\ce{NaOH}$, you would start making a lot of sulfate.

If you wanted to know all of the products and their concentrations, you would have to consider water auto-dissociation as well. The argument I made is not meant to be quantitative, but to show why we don't expect sulfuric acid to be left over. In aqueous solution, there is never much sulfuric acid (even in the absence of $\ce{NaOH}$) because, as a strong acid, it dissociates.