Answer to Question #100011 in General Chemistry for Bobby

Question

How many liters of fluorine gas are needed to form 379 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

p = 2 atm = 202650 Pa = 202650 J/m³

V(SF₆) = 379 l = 0,379 m³

R = 8,314 J/(mol•K)

T = 273,15 K

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pV=nRT

n(SF₆) = PV/RT = 202650•0,379/(8,314•273,15) = 33,82 mol

n(F₂)=3•n(SF₆) = 3•33,82 = 101,46 mol

V(F₂)= nRT/p = 101,46•8,314•273,15/202650 = 1,137 m³ = 1137 liters

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