A 5.00 L tank at 17.5 °C is filled with 19.5 g of boron trifluoride gas and 18.4

Question

A 5.00 L tank at 17.5 °C is filled with 19.5 g of boron trifluoride gas and 18.4 g of carbon monoxide gas. You can assume both under these conditions. 4 g of carbon monoxide gas. You can assume both gases behave as ideal gases 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 boron trifluoride atem partial pressure0 mole fraction: carbon monoxide partial pressure: Total pressure in tank

Explanation / Answer

Molar mass of Boron Triflouride (BF3) = 67.82 g/mol

Number of moles of BF3 = Mass/Molar mass = 19.5/67.82 = 0.2875 moles

Molar mass of Carbon monoxide (CO) = 28.01 g/mol

Number of moles of CO = Mass/Molar mass = 18.4/28.01 = 0.6569 moles

mole fraction of BF3 = numbe rof moles of BF3/(number of moles of BF3+ number of moles of CO)

=> 0.2875/(0.2875+0.6569)

=> 0.3044 = 0.304 (three significant figures)

Using ideal gas equation

PV = nRT

P * 5.00 = 0.2875 * 0.0821 * (273+17.5)

P = 1.37 atm

mole fraction of CO= numbe rof moles of CO/(number of moles of BF3 + number of moles of CO)

=> 0.6569/(0.2875+0.6569)

=> 0.6955 = 0.696 (three significant figures)

Using ideal gas equation

PV = nRT

P * 5.00 = 0.6569 * 0.0821 * (273+17.5)

P = 3.13 atm

Total Pressure = Sum of partial pressure = 1.37 + 3.13 = 4.50 atm

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