How much hydrogen will be released when Na weighing 15.8 interacts with glycerin.

Let’s find the amount of sodium substance.

n = m: M.

M (Na) = 23 g / mol.

n = 15.8 g: 23 g / mol = 0.687 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

2CH2 (OH) – CH (OH) – CH2 (OH) + 6Na → 2CH2 (ONa) _CH (ONa) -CH2 (ONa) + 3H2.

According to the reaction equation, there are 6 mol of sodium for 3 mol of hydrogen. Substances are in quantitative ratios 3: 6 = 1: 2.

The amount of hydrogen will be 2 times less than the amount of sodium.

n (H2) = 1 / 2n (Na) = 0.687: 2 = 0.34 mol.

Let’s find the volume of hydrogen.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 0.34 mol × 22.4 L / mol = 7.616 L.

Answer: 7.616 liters.