In a mixture of hydrogen and methane, the volume fraction of hydrogen is 50%.

In a mixture of hydrogen and methane, the volume fraction of hydrogen is 50%. How many grams of hydrogen is contained in 22.4 liters of the mixture?

If the volume of the mixture is 22.4 liters, and the volume fraction of hydrogen in it is 50%, then the volume of hydrogen will be equal to half the volume of the mixture, 11.2 liters.

To find the mass of hydrogen, you need to multiply its molar mass by the amount of substance:

m = n × M, where M (H2) = 2Ar (H2) = 2 × 1 = 2 g / mol.

If, according to Avogadro’s law, 1 mol occupies a value of 22.4 L / mol, then 11.2 L will be equal to 0.5 mol, or:

1 mol – 22.4 l / mol,

X mol – 11.2 L / mol, hence x = (1 × 11.2 L): 22.4 = 0.5 mol.

We find the mass of hydrogen by the formula m = 2 × 0.5 = 1 g.

Answer: the mass of hydrogen in the mixture is 1 g.