A piece of sodium, weighing 9.2 g, was placed in an excess of water. Determine the mass of the formed alkali

A piece of sodium, weighing 9.2 g, was placed in an excess of water. Determine the mass of the formed alkali. Find the volume of gas evolved.

The equation for the reaction of sodium with water:

2Na + 2H2O = 2NaOH + H2

Let’s find the amount of substance Na:

v (Na) = m (Na) / M (Na) = 9.2 / 23 = 0.4 (mol).

According to the reaction equation, from 2 mol of Na, 2 mol of NaOH and 1 mol of H2 are formed, therefore:

v (NaOH) = v (Na) = 0.4 (mol),

v (H2) = v (Na) / 2 = 0.4 / 2 = 0.2 (mol).

Thus, the mass of the formed alkali:

m (NaOH) = v (NaOH) * M (NaOH) = 0.4 * 40 = 16 (g)

and the volume of gas evolved under normal conditions (n.o.):

V = v (H2) * Vm = 0.2 * 22.4 = 4.48 (l).