What volume in liters of fluorine gas is needed to form 953 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

S(s) + 3F₂(g) ==> SF₆(g) .. balanced equation

Since the temperature and the pressure remain constant, we can simply read the equation as

"3 liters of F₂ produces 1 liter of SF₆

Then use dimensional analysis to find the answer:

953 L SF₆ x 3 L F₂ / 1 L SF6 = 2859 L F₂

The long way:

PV = nRT

Find moles of SF₆

n = PV/RT = (2.00 atm)(953 L) / (0.0821 Latm/Kmol)(273.15K

n = 84.99 mols SF₆

Find moles F₂ needed:

84.99 mols SF₆ x 3 mols F₂ / 1 mol SF₆ = 255 mols F₂

Find volume of F₂ using PV = nRT

V = nRT/P = (255 mols)(0.0821 Latm/Kmol)(273K) / 2.00 atm

V = 2858 L