calculating mole fraction in a gas mixture

Question

A 5.00L

 tank at 24.7°C

 is filled with 13.5g

 of boron trifluoride gas and 12.8g

 of carbon dioxide gas. You can assume both gases behave as ideal gases under these conditions.

Calculate the mole fraction of each gas. Round each of your answers to 

3

 significant digits.

Answer 1

molar mass BF₃ = 67.805 g/mole

molar mass CO₂ = 44.009 g/mole

moles BF₃ present = 13.5 g x 1 mole/67.805 g = 0.1991 moles

moles CO₂ present = 12.8 g x 1 mole/44.009 g = 0.2908 moles

TOTAL moles present = 0.1991 + 0.2908 = 0.4899 moles

Mole fraction BF₃ = 0.1991/0.4899 = 0.406 (to 3 sig. figs.)

Mole fraction CO₂ = 0.2908/0.4899 = 0.594 (to 3 sig. figs.)