2Na (s) + 2H2O (l) → 2NaOH (aq) + H2 (g) When 2 moles of Na react with water at 25°C and 1 atm, the volume of H2 formed is 24.5 L. Calculate the magnitude of work done in joules when 0.45 g of Na reacts with water under the same conditions. (The conversion factor is 1 L · atm = 101.3 J.)

Magnitude of work done = 24.28 J

Explanation:

No. of moles = Reacting mass / Molar mass

Reacting mass of Na = 0.45 g

Molar mass of sodium = 23g/mol

∴ No. of moles of sodium = 0.45/23 = 0.0196 mole

If 2 moles of Na react with water at 25°C and 1 atm, 24.5 L of H₂ was formed.

∴ when 0.0196 mole of Na react with water under the same conditions, (24.5*0.0196) / 2 L of H₂ will be formed.

⇒ (24.5*0.0196) / 2 L = 0.24 L

0.24 L * 1 atm = 0.24 L. atm

Since 1 L · atm = 101.3 J

∴ 0.24 L. atm = (0.24 L. atm * 101.3) / 1 = 24.28 J

Magnitude of work done = 24.28 J