Calculate the volume of hydrogen released when a 4.8 g sodium sample containing 2% impurities interacts with water.

Given:
m tech. (Na) = 4.8 g
ω (approx.) = 2%

To find:
V (H2) -?

1) Write the reaction equation:
2Na + 2H2O => 2NaOH + H2;
2) Calculate the mass fraction of Na:
ω (Na) = 100% – ω (approx.) = 100% – 2% = 98%;
3) Calculate the mass of pure Na:
m clean (Na) = ω (Na) * m tech. (Na) / 100% = 98% * 4.8 / 100% = 4.7 g;
4) Calculate the amount of substance Na:
n (Na) = m (Na) / M (Na) = 4.704 / 23 = 0.2 mol;
5) Determine the amount of substance H2:
n (H2) = n (Na) / 2 = 0.2 / 2 = 0.1 mol;
6) Calculate the volume H2:
V (H2) = n (H2) * Vm = 0.1 * 22.4 = 2.24 liters.

Answer: The volume of H2 is 2.24 liters.