Answer: A volume of 2997 liters of fluorine gas is needed to form 999 L of sulfur hexafluoride gas if the given reaction takes place at 2.00 atm and 273.15 K.
Explanation:
The given reaction is as follows.
[tex]S(s) + 3F_{2}(g) \rightarrow SF_{6}(g)[/tex]
This show that 3 moles of fluorine is reacting to give 1 moles of sulfur hexafluoride.
According to the ideal gas formula,
PV = nRT
This means that volume of a gas is directly proportional to the number of moles. Hence, volume of fluorine required is calculated as follows.
[tex]3 \times 999 L\\= 2997 L[/tex]
Thus, we can conclude that a volume of 2997 liters of fluorine gas is needed to form 999 L of sulfur hexafluoride gas if the given reaction takes place at 2.00 atm and 273.15 K.
