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 dа ta:

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.