Sulfur hexafluoride is produced by reacting elemental sulfur with fluorine gas.S₈(s) + 24 F₂(g) $\to$ 8 SF₆(g)
What is the percent yield if 18.3 g SF₆ is isolated from the reaction of 10.0 g S₈ and 30.0 g F₂?

Question

Sulfur hexafluoride is produced by reacting elemental sulfur with fluorine gas.S₈(s) + 24 F₂(g)

\to

8 SF₆(g)
What is the percent yield if 18.3 g SF₆ is isolated from the reaction of 10.0 g S₈ and 30.0 g F₂?

Quizplus · Accepted Answer

Theoretical yield of SFS1U1B16S1U1B0 can be calculated using stoichiometry as follows:

1 mol SS1U1B18S1U1B0 = 1 mol SFS1U1B16S1U1B0 (according to balanced equation)
Molar mass of SS1U1B18S1U1B0 = 32.1 g/mol
Molar mass of FS1U1B12S1U1B0 = 188.0 g/mol
Therefore, number of moles of SS1U1B18S1U1B0 = 10.0 g / 32.1 g/mol = 0.312 mol
Number of moles of FS1U1B12S1U1B0 = 30.0 g / 188.0 g/mol = 0.159 mol
Limiting reagent is SS1U1B18S1U1B0 (as it produces less number of moles of product)
Number of moles of SFS1U1B16S1U1B0 produced = 0.312 mol
Theoretical yield of SFS1U1B16S1U1B0 = 0.312 mol x  SF1U1B12S1U1B0(g)F1F1F1S1F1F1F10 8 SFS1U1B16S1U1B0(g)
Molar mass of SFS1U1B16S1U1B0 = 146.05 g/mol
Theoretical yield of SFS1U1B16S1U1B0 = 0.312 mol x 146.05 g/mol = 45.49 g
Percent yield = (Actual yield / Theoretical yield) x 100
Percent yield = (18.3 g / 45.49 g) x 100 = 40.2%

Sulfur Hexafluoride Is Produced by Reacting Elemental Sulfur with Fluorine →\to→

Sulfur Hexafluoride Is Produced by Reacting Elemental Sulfur with Fluorine $\to$