Limiting reactant and percent yield
The first step is to figure out a reaction equation.

[I'm assuming elemental] sodium reacts with water violently because it releases a lot of energy when its only valence electron gets ripped apart. It creates \ce{NaOH} and \ce{H2}.

$\ce{Na + H2O -&gt; NaOH + H2}$

We need to balance this.

$\ce{2Na + 2H2O -&gt; 2NaOH + H2}$

$\ce{Na}$ and $\ce{H2O}$ are consumed at equal rates, so the $1.95$ moles of $\ce{H2O}$ are going to be used up completely (leaving $2.30 - 1.95 = 0.35$ moles of $\ce{Na}$ not reacted).

$\ce{NaOH}$ is formed at the same rate of consumption of the other two, so it's going to have $1.95$ moles produced.

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Mathematically...

$\frac{2.30mol_\ce{Na}}{2mol_\ce{Na}} = 1.15$

and

$\frac{1.95mol_\ce{water}}{2mol_\ce{water}} = 0.975$

Effectively, water burns out faster in the reaction than sodium. So, we have to base our calculations off of moles of water.

For every 2 moles of water used, 2 moles of sodium hydroxide are formed.

$1.95mol_\ce{water} * \frac{2mol_\ce{NaOH}}{2mol_\ce{water}} = 1.95mol_\ce{NaOH}$