Sulfur and fluorine react in a combination reaction to produce sulfur hexafluoride: S (s) + 3F₂ (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 find the maximum amount of SF₆ produced, determine the limiting reactant. Molar mass of S = 32.07 g/mol, F₂ = 38.00 g/mol, SF₆ = 146.07 g/mol. Moles of S = 3.5 g / 32.07 g/mol = 0.109 moles. Moles of F₂ = 4.5 g / 38.00 g/mol = 0.118 moles. Reaction requires 3 moles of F₂ per mole of S, so 0.109 moles S needs 0.327 moles F₂. F₂ is limiting. Moles of SF₆ = 0.118 moles F₂ / 3 = 0.0393 moles. Mass of SF₆ = 0.0393 moles * 146.07 g/mol = 5.8 g.

Sulfur and Fluorine React in a Combination Reaction to Produce