26 August, 09:15

# A gas mixture contains 7.0 g N2, 2.0 g H2, and 16.0 g CH4. What is the mole fraction of H2 in the mixture? (c) Calculate the pressure of the gas mixture and the partial pressure of each constituent gas if the mixture is in a 5.0 L vessel at 20°C.

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1. 26 August, 10:49
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The mole fraction of H₂ gas in the given mixture: χ₁ = 0.44

The total pressure of the gas mixture: P = 10.825 atm

The partial pressure of H₂ gas in the given mixture: p₁ = 4.76 atm

The partial pressure of N₂ gas in the given mixture: p₂ = 1.19 atm

The partial pressure of CH₄ gas in the given mixture: p₃ = 4.76 atm

Explanation:

Given: mass of H₂: W₁ = 2 g, mass of N₂: W₂ = 7 g, mass of CH₄: W₃ = 16 g

Molar mass of H₂: M₁ = 2g/mol, molar mass of N₂: M₂ = 28g/mol, molar mass of CH₄: M₃=16g/mol

Number of moles of H₂: n₁ = W₁ : M₁ = 2 : 2 = 1 mol

Number of moles of N₂: n₂ = W₂ : M₂ = 7 : 28 = 0.25 mol

Number of moles of CH₄: n₃ = W₃ : M₃ = 16 : 16 = 1 mol

Therefore, the total number of moles: n = n₁ + n₂ + n₃ = 1 + 0.25 + 1 = 2.25 mol

The mole fraction of H₂ gas in the given mixture: χ₁ = n₁ : n = 1 : 2.25 = 0.44

The mole fraction of N₂ gas in the given mixture: χ₂ = n₂ : n = 0.25 : 2.25 = 0.11

The mole fraction of CH₄ gas in the given mixture: χ₃ = n₃ : n = 1 : 2.25 = 0.44

The total pressure of the gas mixture can be calculated by the ideal gas equation: P = n. R. T : V

Given: total volume:V = 5L, total number of moles:n = 2.25 mol, gas constant: R = 0.08206 L. atm. mol⁻. K⁻, temperature: T = 20°C = 20 + 273.15 = 293.15 K, total pressure: P=?

Therefore, total pressure of the gas mixture: P = (2.25 mol * 0.08206 L. atm. mol⁻. K⁻ * 293.15 K) : (5L) = 10.825 atm

The partial pressure of H₂ gas in the given mixture: p₁ = χ₁ * P = 0.44 * 10.825 atm = 4.76 atm

The partial pressure of N₂ gas in the given mixture: p₂ = χ₂ * P = 0.11 * 10.825 atm = 1.19 atm

The partial pressure of CH₄ gas in the given mixture: p₃ = χ₃ * P = 0.44 * 10.825 atm = 4.76 atm