How many liters of fluorine gas are needed to form 879 l of sulfur hexafluoride gas if the following reaction takes place at 2.00 atm and 273.15 k

2637.3 litters of fluorine gas are needed to produce 879 litters of sulfur hexafluoride.

Explanation:

Given data:

Volume of SF₆ = 879 L

Pressure = 2 atm

Temperature = 273.15 k

Volume of fluorine required = ?

Solution:

Balance chemical equation:

    S (s) + F₂(g)    →  SF₆(g)

First of all we will calculate the moles of SF₆.

  PV = nRT

  n = PV/RT

  n = 2. atm× 879L / 0.0821 L. atm. mol⁻¹. K⁻¹  × 273.15 K

  n = 1758 atm. L/ 22.43 L. atm. mol⁻¹

  n = 78.4 mol

78.4 moles of SF₆ will produce.

Now we will compare the moles of SF₆ and fluorine from balance chemical equation.

      SF₆   :   F

        1     :    3

      78.4  :   3/1 × 78.4 = 235.2 moles

Now we will calculate the volume of fluorine.

   PV = nRT

  V  = nRT / P

  V= 235.2 mol. 0.0821 L. atm. mol⁻¹. K⁻¹  × 273.15 K / 2 atm

  V = 5274. 5 / 2

  V = 2637.3 L

2637.3 litters of fluorine gas are needed to produce 879 litters of sulfur hexafluoride.

the answer is na (sodium)