When sodium interacted with an amount of 0.5 mol of substance with water, hydrogen was obtained

When sodium interacted with an amount of 0.5 mol of substance with water, hydrogen was obtained with a volume of 4.2 L. Calculate the practical gas yield.

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

2Na + 2H2O = 2NaOH + H2.

According to the reaction equation, there is 1 mole of hydrogen per 2 mol of sodium. Substances are in quantitative ratios of 2: 1.

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

2 mol Na – 1 mol H2

0.5 mol Na – x mol H2

x mol H2 = (0.5 × 1): 2 = 0.25 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 = 25 mol × 22.4 L / mol = 5.6 L.

5.6 liters should be obtained theoretically (according to calculations).

According to the condition of the problem, it turned out 4.2 liters.

Let us find the volumetric yield of hydrogen from the theoretically possible one.

5.6 l – 100%,

4.2 – x%,

X = (4.2 × 100): 5.6 L = 75%.

Or by the formula:

Ɣ = Vpractical: V theoretical × 100%.

Ɣ = (4.2 × 100): 5.6 l = 75%.

Answer: 75%.