Question 12 A tank at is filled with of boron trifluoride gas and of sulfur tetrafluoride 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 significant digits.

BF₃ ⇒ 0.891

SF₄ ⇒ 0.109

Explanation:

It seems the question is incomplete, in that case I will answer using the values that have been found in simmilar questions online. If the values in your problem vary, keep that in mind when solving it.

" A 9.00 L tank at 23.6 °C is filled with 12.2 g of boron trifluoride gas and 2.37 g of sulfur tetrafluoride 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. "

First we convert g of both compounds to mol, using their molar mass:

Boron trifluoride BF₃ ⇒ Molar mass = 67,82 g/mol

12.2 g : 67.82 g/mol = 0.180 mol BF₃

Sulfur tetrafluoride SF₄ ⇒ Molar mass = 108 g/mol

2.37 g : 108 g/mol = 0.022 mol SF₄

Total number of moles = moles BF₃ + moles SF₄ = 0.180 + 0.022 = 0.202

Finally we calculate the mole fracion of each gas:

BF₃ ⇒ 0.180 / 0.202 = 0.891 SF₄ ⇒ 0.022 / 0.202 = 0.109