How much hydrogen will be released when 40 g of glycerin interacts with 40 g of metallic sodium?

In accordance with the condition of the problem, we write the equation:
2С3Н5 (ОН) 3 + 6Na = 3H2 + 2CH2 – Ona – CHONa – CH2 – Ona – substitution reaction, hydrogen and sodium glycerate are released;
Let’s make calculations using the formulas:
M C3H5 (OH) 3 = 92 g / mol;
M (Na) = 22.9 g / mol.
Determine the amount of moles of glycerin, sodium, if the mass is known:
Y C3H5 (OH) 3 = m / M = 40/92 = 0.43 mol (deficient substance);
Y (Na) = m / M = 40 / 22.9 = 1.75 mol (substance in excess);
In carrying out the calculations, we take into account the substance in the deficiency.
We make the proportion:
0.43 mol of glycerol – X mol (H2);
-2 mol -3 mol from here, X mol (H2) = 0.43 * 3/2 = 0.645 mol.
Find the volume of hydrogen:
V (H2) = 0.645 * 22.4 = 14.45 l.
Answer: the volume of hydrogen is 14.45 liters.