Write the equation for the reaction of ethanol with sodium. Determine the amount of hydrogen that will be released

Write the equation for the reaction of ethanol with sodium. Determine the amount of hydrogen that will be released by the action of sodium on 27.6 grams of ethanol, if the product yield is 85% of the theoretically possible.

The equation for the reaction of ethanol with sodium:

2C2H5OH + 2Na = 2C2H5ONa + H2

Amount of C2H5OH substance:

v (C2H5OH) = m (C2H5OH) / M (C2H5OH) = 27.6 / 46 = 0.6 (mol).

According to the reaction equation, 1 mol of H2 is formed per 2 mol of C2H5OH, therefore, taking into account the product yield:

v (H2) = (v (C2H5OH) / 2) * 0.85 = (0.6 / 2) * 0.85 = 0.255 (mol).

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

V (H2) = v (H2) * Vm = 0.255 * 22.4 = 5.712 = 5.7 (l).