Answer:
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.
