If 4.00 moles of oxygen gas, 3.00 moles of hydrogen gas, and 1.00 mole of nitrogen gas are combined in a closed container at standard pressure, what is the partial pressure exerted by the hydrogen gas?

The partial pressure exerted by the hydrogen gas, H₂ in the container is 0.375 atm

We'll begin by calculating the mole fraction of H₂ in the gas mixture. this can be obtained as follow:

Mole of O₂ = 4 moles

Mole of H₂ = 3 moles

Mole of N₂ = 1 mole

Total mole = 4 + 3 + 1 = 8 moles

Mole fraction of H₂ = 0.375

Total pressure = STP = 1 atm

Partial pressure = mole fraction × Total Pressure

Partial pressure of H₂ = 0.375 × 1

Therefore, the partial pressure of H₂ in the container is 0.375 atm

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Partial pressure of H₂ is 0.375 atm

Explanation:

We apply the mole fraction to solve this.

Standard pressure is 1 atm

Mole fraction of a gas = Moles of gas / Total moles

Mole fraction of pressure = Partial pressure of gas / Total pressure

Both values are the same

Total moles = 4 moles of O₂  +  3 moles of H₂ and 1 mol of N₂ = 8 moles

3 moles H₂ / 8 moles = Partial pressure H₂ / 1 atm

(3 / 8 ) .1 = 0.375  atm → Partial pressure of H₂

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