Calculating partial pressure in a gas mixture A 10.00 L tank at 3.43 °C is fille

Question

Calculating partial pressure in a gas mixture A 10.00 L tank at 3.43 °C is filled with 19.0 g of chlorine pentafluoride gas and 12.2 g of sulfur hexafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calcula digits. te 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 mole fraction: chlorine pentafluoride partial pressure: mole fraction: sulfur hexafluoride partial pressure: atam Total pressure in tank:

Explanation / Answer

Molar mass of Chlorine Pentaflouride (ClF5) = 130.445 g/mol

Number of moles of ClF5 = Mass/Molar mass = 19.0/130.445 = 0.1456 moles

Molar mass of Sulphur hexaflouride (SF6) = 146.06 g/mol

Number of moles of SF6 = Mass/Molar mass = 12.2/146.06 = 0.083 moles

mole fraction of ClF5 = numbe rof moles of ClF5/(number of moles of ClF5 + number of moles of SF6)

=> 0.1456/(0.1456+0.083)

=> 0.6369 = 0.637 (three significant figures)

Using ideal gas equation

PV = nRT

P * 10.00 = 0.1456 * 0.0821 * (273+3.43)

P = 0.330 atm

mole fraction of SF6 = numbe rof moles of SF6/(number of moles of ClF5 + number of moles of SF6)

=> 0.083/(0.1456+0.083)

=> 0.3630 = 0.363 (three significant figures)

Using ideal gas equation

PV = nRT

P * 10.00 = 0.083 * 0.0821 * (273+3.43)

P = 0.188 atm

Total Pressure = Sum of partial pressure = 0.330 + 0.188 = 0.518 atm

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