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A gas phase mixture of H2 and N2 has a total pressure of 784 torr with an H2 partial pressure of 124 torr. What mass of N2 gas is present in 2.00 L of the mixture at 298 K

Answer :

EDUARD22SLYIN

The mass of N₂ in the mixture having a total pressure of 784 torr with an H₂ partial pressure of 124 torr is 1.988 g

We'll begin by calculating the partial pressure of N₂. This can be obtained as follow:

Total pressure = 784 torr

Partial pressure of H₂ = 124 torr

Partial pressure of N₂ =?

Total pressure = Partial pressure of H₂ + Partial pressure of N₂

784 = 124 + Partial pressure of N₂

Collect like terms

Partial pressure of N₂ = 784 – 124

Partial pressure of N₂ = 660 Torr

  • Next, we shall determine the number of mole of N₂ in the mixture.

Pressure of N₂ (P) = 660 Torr

Volume (V) = 2 L

Temperature (T) = 298 K

Gas constant (R) = 62.364 L.Torr/Kmol

Number of mole of N₂ (n) =?

PV = nRT

660 × 2 = n × 62.364 × 298

1320 = n × 18584.472

Divide both side by 18584.472

n = 1320 / 18584.472

n = 0.071 mole

  • Finally, we shall determine the mass of N₂ in the mixture.

Mole of N₂ = 0.071 mole

Molar mass of N₂ = 2 × 14 = 28 g/mol

Mass of N₂ =?

Mass = mole × molar mass

Mass of N₂ = 0.071 × 28

Mass of N₂ = 1.988 g

Therefore, the mass of N₂ in the mixture having a total pressure of 784 torr with an H₂ partial pressure of 124 torr is 1.988 g

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