How many liters of fluorine gas are needed to
form 919 L of sulfur hexafluoride gas if the
following reaction takes place at 2.00 atm
and 273.15 K:
S(s) + 3F₂(g) → SF. (g)?

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tripatsingh039 tripatsingh039 · Answer 1 · 2022-07-26T10:37:53+02:00

2756.1 liters of fluorine gas is needed to produce 919 liters of sulfur hexafluoride.

The volume of SF₆ = 919 L

Pressure = 2 atm

Temperature = 273.15 k

The volume of fluorine required =?

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× 919L / 0.0821 L. atm. mol⁻¹. K⁻¹ × 273.15 K

n = 1838 atm. L/ 22.43 L. atm. mol⁻¹

n = 81.9 mol

81.9 moles of SF₆ will produce.

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

SF₆: F

1 : 3

81.9 : 3/1 × 81.9 = 245.7 moles

Now we will calculate the volume of fluorine.

PV = nRT

V = nRT / P

V= 245.7 mol × 0.0821 L. atm. mol⁻¹. K⁻¹ × 273.15 K / 2 atm

V = 5512.2 / 2

V = 2756.1 L

2756.1 liters of fluorine gas are needed to produce 919 liters of sulfur hexafluoride.

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