Calculating partial pressure in a gas mixture A 6.00 L tank at 25.4 °C is filled with 8.90 g of carbon monoxide gas and 3.63 g of chlorine pentafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Be sure your answers have the correct number of significant digits.
Mole fraction: carbon monoxide partial pressure atm mole fraction chlorine pentafluoride partial pressure atm Total pressure in tank: atm

Hydrogen, electrocatalysts, pure metals, anhydrides, formic, methyl formate, N,N-dimethylformamide, benzoic acid, and as a reduction agent in coke ovens are all produced using carbon monoxide.

Briefing:

We should apply the Ideal Gases Law to solve this:

P . V = n . R . T

We need n, which is the total moles for the mixture

Total moles = Moles of CO + Moles of ClF₅

Moles of CO = mass of CO / molar mass CO → 8.9 g/28 g/mol = 0.317 mol

Moles of ClF₅ = mass of ClF₅ / molar mass ClF₅ → 3.63g/ 130.45 g/m = 0.0278 mol

0.317 mol + 0.0278 mol → 0.398 moles in the mixture

So we have the total moles so with the formula we would know the total pressure.

P . 6L = 0.724 mol . 0.082L.atm/mol.K . 292.2K

P = ( 0.724 mol . 0.082L.atm/mol.K . 292.2K) / 6L

P = 2.89 atm

Mole fraction is defined as the quotient between the moles of gas over total moles, and it is equal to partial pressure of that gas over total pressure

Partial pressure for gas X/Total pressures = Moles of gases X/Total mole

(Moles of gas X / Total moles) . Total pressure = Partial pressure of gas X

Mole fraction CO = 0.317 / 0.724 = 0.437

Partial pressure CO = 0.88 . 2.89 atm → 2.56 atm

Mole fraction ClF₅ = 0.0278 / 0.724 = 0.0383

Partial pressure ClF₅ = 0.0383 . 2.89 atm → 0.110 atm

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