Mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm. Calculate the mass percent of these two gases. Assume ideal-gas behavior.

The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant =

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

Equating the moles of the individual gases to the moles of mixture:

To calculate the mass percentage of substance in mixture we use the equation:

Mass of the mixture = 13.22 g

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

in chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. the standard formula is c = m/v, where c is the concentration, m is the mass of the solute dissolved, and v is the total volume

your answer would be “dry”