1. Write the balanced chemical equation:

2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)

2. Calculate the moles of sodium (Na):

* Molar mass of Na = 22.99 g/mol

* Moles of Na = (1.20 g) / (22.99 g/mol) = 0.0522 mol

3. Determine the limiting reactant:

* Water:

* Density of water = 1 g/mL

* Mass of water = (3.00 mL) * (1 g/mL) = 3.00 g

* Molar mass of water = 18.015 g/mol

* Moles of water = (3.00 g) / (18.015 g/mol) = 0.166 mol

* Ratio: The balanced equation shows a 2:2 ratio between Na and H₂O. This means 0.0522 mol of Na would require 0.0522 mol of H₂O to react completely. Since we have more water (0.166 mol) than needed, sodium is the limiting reactant.

4. Calculate the moles of NaOH produced:

* The balanced equation shows a 2:2 ratio between Na and NaOH.

* Moles of NaOH produced = 0.0522 mol Na * (2 mol NaOH / 2 mol Na) = 0.0522 mol NaOH

5. Calculate the mass of NaOH produced:

* Molar mass of NaOH = 39.997 g/mol

* Mass of NaOH produced = (0.0522 mol) * (39.997 g/mol) = 2.09 g NaOH

Therefore, approximately 2.09 grams of NaOH will be produced.

Important Note: This reaction is highly exothermic and produces hydrogen gas, which is flammable. It should only be performed under controlled conditions by a trained professional.