Answer to Question #142603 in General Chemistry for mitch julmice

Question

How many liters of fluorine gas are needed to form 999 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)

........... ?L ...........999L ................ at 2.00 atm and 273.15K

Thanks to Gay-Lussac (law of combining volumes) and Avogadro we know that the volumes of gases in a reaction are in the same ratio as the coefficients in the balanced chemical equation.

999L SF6 x (3 L F2 / 1 L SF6) = 2997 L F2.... round to 3 significant digits: 3.00x10^3L

Of course you could crank it through the ideal gas equation .... and get the same result, but that's a lot of extra work. The following is an example of what not to do. :)

PV = nRT

n = PV / RT

n = 2.00 atm x 999L / 0.08206 Latm/molK / 273.15K

n = 89.14 mol SF6

89.14 mol SF6 x (3 mol F2 / 1 mol SF6) = 267.4 mol F2

PV = nRT

V = nRT / P

V = 267.4 mol x 0.08206 Latm x 273.15K / 2.00 atm

V = 2997L .......... round to three significant digits: 3.00x10^3L F2