A 6.000L tank at 19.2°C is filled with 18.0g of carbon monoxide gas and 10.6g 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. Round each of your answers to 3 significant digits.

Question

Chemistry

Ali Bates

30 July, 08:23

A 6.000L tank at 19.2°C is filled with 18.0g of carbon monoxide gas and 10.6g 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. Round each of your answers to 3 significant digits.

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Pookie · Answer 1

Total pressure: 2.89 atm

Mole fraction CO: 0.88

Partial pressure CO: 2.56 atm

Mole fraction ClF₅: 0.12

Partial pressure ClF₅: 0.33 atm

Explanation:

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 → 18 g/28 g/mol = 0.643 mol

Moles of ClF₅ = mass of ClF₅ / molar mass ClF₅ → 10.6g / 130.45 g/m = 0.0812 mol

0.643 mol + 0.0812 mol → 0.724 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

Moles of gas X / Total moles = Partial pressure of gas X/Total pressure

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

Mole fraction CO = 0.643 / 0.724 = 0.88

Partial pressure CO = 0.88. 2.89 atm → 2.56 atm

Mole fraction ClF₅ = 0.0812 / 0.724 = 0.12

Partial pressure ClF₅ = 0.12. 2.89 atm → 0.33 atm