Calculation of the volume of HCl required for neutralization

To determine the volume of HCl required for the exact neutralization of the sodium hydroxide (NaOH) produced, we need to consider the stoichiometry of the reaction and the molar ratios involved.

1. Balanced chemical equation
The balanced chemical equation for the reaction between sodium (Na) and water (H2O) to produce sodium hydroxide (NaOH) is as follows:

2Na + 2H2O -> 2NaOH + H2

This equation tells us that for every 2 moles of sodium (Na) reacting, 2 moles of sodium hydroxide (NaOH) are produced. Therefore, the molar ratio between Na and NaOH is 1:1.

2. Molar mass of sodium (Na)
The molar mass of sodium (Na) is approximately 23 g/mol.

3. Calculation of moles of sodium (Na)
Given that the mass of sodium (Na) is 0.46 g, we can calculate the number of moles using the formula:

moles = mass / molar mass

moles = 0.46 g / 23 g/mol

moles = 0.02 mol

4. Molar ratio between Na and NaOH
Since the molar ratio between Na and NaOH is 1:1, the number of moles of sodium hydroxide (NaOH) produced will also be 0.02 mol.

5. Molarity of HCl
To calculate the volume of HCl required, we need to know the molarity of the HCl solution. Let's assume the molarity is given as 1 M (1 mol/L).

6. Calculation of volume of HCl
Using the molar ratio between NaOH and HCl, which is also 1:1, we can determine the volume of HCl required using the formula:

moles = volume (L) x molarity

0.02 mol = volume (L) x 1 mol/L

volume (L) = 0.02 L

Since 1 L is equal to 1000 mL, the volume of HCl required is 20 mL.

Conclusion

To exactly neutralize the amount of NaOH produced from the reaction of 0.46 g of sodium with water, approximately 20 mL of HCl solution with a molarity of 1 M would be sufficient. It is important to note that this calculation assumes ideal conditions and complete reaction conversion.