What is the maximum volume of hydrogen that will displace metallic sodium from glycerin weighing 3.2 g?

According to the condition of the problem, we compose an equation, select the coefficients:
2С3Н5 (ОН) 3 + 6Na = 3H2 + 2CH2O – Na – CHO – Na – CH2O – Na – substitution reaction, hydrogen gas and sodium glycerate are released;
Let’s calculate the molar masses of glycerin and hydrogen:
M C3H5 (OH) 3 = 92 g / mol;
M (H2) = 2 g / mol;
Determine the amount of moles of glycerin by the formula:
Y C3H5 (OH) 3 = m / M = 3.2 / 92 = 0.03 mol.
Let’s make the proportion:
0.03 mol C3H5 (OH) 3 – X mol (H2);
-2 mol -3 mol from here, X mol (H2) = 0.03 * 3/2 = 0.045 mol.
Find the volume of hydrogen:
V (H2) = 0.045 * 22.4 = 1 liter.
Answer: 1 liter of hydrogen is released.