What volume in liters of fluorine gas is needed to form 687 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

Hello, Mae,

I'll solve this using two approaches. The first is longer, but demonstrates the underlying calculations. The second is much shorter, but depends on the fact that the conditions for both the reactant and product gasses are the same.

Use the Ideal Gas Law

687 liters of sulfur hexafluoride, SF₆, is a lot of gas, when you consider it is at 2 atm. 1 mole of any gas occupies 22.4 liters at STP. Temperature here is standard temperature, but the pressure is twice standard. This means we'd have twice that volume at 1 atm. The ideal gas law, PV = nRT, can be used to calculate the number of moles of sulfur hexafluoride in 687 liters under these conditions.

n = (PV)/(RT)

n = (2 atm)(687L)/(0.08206 L*atm*K⁻¹⋅mol⁻¹)(273.15K)

n = 61.3 moles SF₆

The balanced equation tells us we need 3 moles F₂ per mole SF₆. So we need 3 (moles F₂/mole SF₆)*(61.3 moles SF₆) = 183.9 moles F₂. We can use the ideal gas law again to find the volume of F₂ under the same conditions.

V = nRT/P

V = 2061 liters F₂ at 2 atm and 273.15K.

Skip The Line Ideal Gas Law

There is actually a much quicker solution to the problem. We know that 1 mole of all gases occupy 22.4 liters at STP. If the problem had been at STP, we could have simply multiplied the 687 liters of SF₆ by 3 to find the volume of F₂ required, since the volumes of the gases are equivalent directly proportional to their volumes.

Athought we are not at STP here, the volumes are still proportional. So however many moles of SF₆ are in 687 liters at non-STP, the 3 to 1 proportion for the F₂ is still valid for the same conditions. So 3*687 liters = 2061 liters at 2 atm is a valid, and much more fun, approach that gives us the same answer.

Bob