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Answer :
Final answer:
After converting the given masses to moles and determining the limiting reactant, we found that fluorine is the limiting reactant. A total of 20.5 g of sulfur remains unreacted, which is answer choice (a).
Explanation:
The subject of this question is Chemistry, specifically related to stoichiometry, which is a branch of chemistry that deals with the quantitative relationships of the substances as they participate in chemical reactions. The student's question asks about determining the mass of excess reactant remaining after a reaction between sulfur (S) and fluorine (F₂) to form sulfur hexafluoride (SF₆).
To solve this problem, we will need to calculate which reactant is the limiting reactant and how much of the other reactant is left unreacted. The balanced chemical equation for the reaction is S(s) + 3 F₂(g) → SF₆(g). We must first convert the given masses of sulfur and fluorine into moles, using their molar masses (S = 32.07 g/mol and F₂ = 38.00 g/mol).
For sulfur:
50.0 g S × (1 mol S / 32.07 g S) = 1.560 moles S
For fluorine:
105.0 g F₂ × (1 mol F₂ / 38.00 g F₂) = 2.763 moles F₂
According to the balanced equation, 1 mole of S reacts with 3 moles of F₂. Therefore, we have excess fluorine because 1.560 moles of sulfur would require 3 × 1.560 = 4.680 moles of fluorine to completely react, and we only have 2.763 moles of fluorine. Hence, fluorine is the limiting reactant.
To find the excess mass of sulfur, we need to calculate how much sulfur reacted with the available fluorine. Since 2.763 moles of F₂ fully react with 2.763 ÷ 3 = 0.921 moles of S:
0.921 moles S × 32.07 g/mol = 29.53 g S reacted
So, the initial mass of sulfur was 50.0 g and 29.53 g reacted, leaving:
50.0 g - 29.53 g = 20.47 g S
Therefore, 20.5 g S of sulfur remains unreacted, which is closest to answer choice (a).
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