Question

How many liters of fluorine gas are needed to form 701 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

To form 701 L of sulfur hexafluoride gas, 2103 liters of fluorine gas are needed.

Step-by-step explanation:

To determine the number of liters of fluorine gas needed to form 701 L of sulfur hexafluoride gas, we can use the stoichiometry of the reaction:

S(s) + 3F₂(g) → SF6(g)

From the balanced equation, we know that 3 moles of fluorine gas are required to react with 1 mole of sulfur hexafluoride. To convert from moles to liters, we can use the ideal gas law:

1 mole = 22.4 L at STP (Standard Temperature and Pressure)

Therefore, to find the number of liters of fluorine gas needed, we can set up the following proportion:

(3 moles F₂/1 mole SF6) = (x liters F₂/701 L SF6)

By cross-multiplying, we find that:

x liters F₂ = (3 moles F₂/1 mole SF6) * 701 L SF6

x liters F₂ = 2103 L F₂

So, 2103 liters of fluorine gas are needed to form 701 L of sulfur hexafluoride gas.