Indicate the mass and volume of hydrogen released when interacting with sodium water weighing 27.6 g.

The equation for the reaction of sodium with water:

2Na + 2H2O = 2NaOH + H2

The amount of sodium substance:

v (Na) = m (Na) / M (Na) = 27.6 / 23 = 1.2 (mol).

According to the reaction equation, 1 mol of H2 is formed per 2 mol of Na, therefore:

v (H2) = v (Na) / 2 = 1.2 / 2 = 0.6 (mol).

Thus, the mass of the released hydrogen is:

m (H2) = v (H2) * M (H2) = 0.6 * 2 = 1.2 (g)

and the amount of hydrogen evolved, measured under normal conditions (n.o.):

V (H2) = v (H2) * Vm = 0.6 * 22.4 = 13.44 (l).