Question

What volume in liters of fluorine gas is needed to form 999 L of sulfur hexafluoride gas if the following reaction takes place at 2.00 atm and 273.15 K: S(s) + 3F₂(g) → SF₆ (g)

Answer 1

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.

Step-by-step explanation:

The given reaction is as follows.

`S(s) + 3F_(2)(g) \rightarrow SF_(6)(g)`

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.

`3 * 999 L\\= 2997 L`

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.