Sodium weighing 2.3 g was placed in ethanol. How much hydrogen is released?

Let’s write the reaction equation:

2C2H5OH + 2Na = 2C2H5ONa + H2

Let’s find the amount of sodium substance:

v (Na) = m (Na) / M (Na) = 2.3 / 23 = 0.1 (mol).

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

v (H2) = v (Na) / 2 = 0.1 / 2 = 0.05 (mol).

Thus, the volume of released hydrogen, measured under normal conditions (n.o.):

V (H2) = v (H2) * Vm = 0.05 * 22.4 = 1.12 (l).

Answer: 1.12 l.