Sulfur and fluorine react in a combination reaction to produce sulfur hexafluoride: S (s) + 3 F₃ (g) → SF₆ (g)
The maximum amount of SF₆ that can be produced from the reaction of 3.5 g of sulfur with 4.5 g of fluorine is __________ g.

Question

Quizplus · Accepted Answer

To determine the maximum amount of sulfur hexafluoride that can be produced, we need to first determine which reactant is limiting. 
- Calculate the number of moles of sulfur: 
3.5 g / 32.06 g/mol = 0.109 mol 
- Calculate the number of moles of fluorine: 
4.5 g / 38.00 g/mol = 0.118 mol 
Since there are 3 moles of fluorine required for every 1 mole of sulfur, we can see that sulfur is the limiting reactant since there are fewer moles of sulfur than fluorine. 
Next, we need to determine the theoretical yield of sulfur hexafluoride: 
- Calculate the number of moles of sulfur hexafluoride that can be produced from the 0.109 moles of sulfur: 
0.109 mol sulfur × (1 mol sulfur hexafluoride / 1 mol sulfur) × (1 mol fluorine / 3 mol sulfur hexafluoride) = 0.0363 mol sulfur hexafluoride 
- Calculate the mass of sulfur hexafluoride that can be produced from the 0.0363 moles of sulfur hexafluoride: 
0.0363 mol sulfur hexafluoride × 146.06 g/mol = 5.29 g sulfur hexafluoride 
Therefore, the maximum amount of sulfur hexafluoride that can be produced from the reaction of 3.5 g of sulfur with 4.5 g of fluorine is 5.29 g. The closest answer choice is C) 5.8 g.

Sulfur and Fluorine React in a Combination Reaction to Produce