Question

How many liters of fluorine gas are needed to form 295 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

Given temperature (T) = 273.15 K

Given pressure (P) = 2.00 atm

Given Volume of SF6 (V) = 295 L

Gas constant (R) = 0.082 L atm/K/mol

Using ideal gas equation, i.e. PV = nRT number of moles of SF6 (n) can be calculated as

n = PV/RT

So,

n= (2.00 atm × 295 L) / (0.082 L atm/K/mol × 273.15 K)

= 26.34 moles

It means moles of SF6 produced = 26.34 moles

Chemical reaction is:

S(s) + 3F2(g) SF6(g)

Now, from balanced chemical equation, it is clear that

1 moles of SF6 requires amount of F2 for production = 3 moles

So, 26.34 moles of SF6 will require amount of F2 for production = 3 × 26.34 = 79.02 moles

It means moles of F2 (n) = 79.02 moles

Temperature (T) = 273.15 K

Pressure (P) = 2.00 atm

Gas constant (R) = 0.082 L atm/mol/K

Volume (V) = ?

So, using ideal gas equation,

V = nRT/P

V = (79.02 moles × 0.082 L atm/mol/K × 273.15 K) / 2 atm

= 884.96 L

Volume of flourine gas required is 884.96 L