P4(s) + 10 F2(g) → 4 PF5(g) How many moles of fluorine gas are required to form 100.0 g of phosphorus pentafluoride?
- az_lenderLv 72 years ago
The molar mass of PF5 is 125.966, so 100.0 grams are 0.79386 moles.
For each mole of PF5 produced, one needs (10/4) mole of fluorine gas, or 2.5 moles of fluorine gas.
Therefore the answer to the question is
(0.79386 moles)*(2.5) = 1.985 moles of fluorine gas.