Answer :
The reaction between 5.00 mol S₈ and 5.00 mol F₂ will produce 20.0 mol of sulfur hexafluoride (SF₆).
In the balanced chemical equation for the reaction between S₈ and F₂, we have:
S₈ + 24F₂ → 8SF₆
The stoichiometric ratio between S₈ and SF₆ is 1:8. This means that for every 1 mol of S₈ reacted, 8 mol of SF₆ are produced. Since we have 5.00 mol of S₈, we can calculate the amount of SF₆ formed as follows:
5.00 mol S₈ × (8 mol SF₆ / 1 mol S₈) = 40.0 mol SF₆
However, we need to consider that we have 5.00 mol of F₂ as well. Since the stoichiometric ratio between F₂ and SF₆ is 24:8, we can calculate the limiting reagent by comparing the amount of F₂ available with the stoichiometric ratio:
5.00 mol F₂ × (8 mol SF₆ / 24 mol F₂) = 1.67 mol SF₆
Since the stoichiometric ratio shows that we can only form 1.67 mol of SF₆ with the available 5.00 mol of F₂, we conclude that F₂ is the limiting reagent. Therefore, the amount of SF₆ formed will be limited by the 1.67 mol of F₂, and the remaining S₈ will not react completely.
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