Calculate the mass of sodium hydroxide produced from 2.3 g of sodium reacting with water.

The atomic mass of Na is given as 23.

$$\text{Molar mass of Na} = 23 \, \text{g/mol}$$

Given mass of Na = $2.3 \, \text{g}$.

$$\text{Moles of Na} = \frac{\text{Mass of Na}}{\text{Molar mass of Na}} = \frac{2.3 \, \text{g}}{23 \, \text{g/mol}} = 0.1 \, \text{mol}$$

The balanced equation is: $2\text{Na(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2\text{(g)}$
From the equation, 2 moles of Na produce 2 moles of NaOH.
The mole ratio of Na : NaOH is $2:2$, which simplifies to $1:1$.
Therefore, moles of NaOH produced = moles of Na reacted.

$$\text{Moles of NaOH} = 0.1 \, \text{mol}$$

Atomic masses: H = 1, O = 16, Na = 23.

$$\text{Molar mass of NaOH} = \text{Na} + \text{O} + \text{H} = 23 + 16 + 1 = 40 \, \text{g/mol}$$

$$\text{Mass of NaOH} = \text{Moles of NaOH} \times \text{Molar mass of NaOH}$$ $$\text{Mass of NaOH} = 0.1 \, \text{mol} \times 40 \, \text{g/mol} = 4.0 \, \text{g}$$

The calculated mass of sodium hydroxide is $4.0 \, \text{g}$.

Comparing this to the given options:
A 0.40g
B 0.80g
C 4.00g
D 8.00g

The correct option is C.

The final answer is $\boxed{\text{C}}$