Answer:
BF₃ ⇒ 0.891
SF₄ ⇒ 0.109
Step-by-step 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