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Answer :
Final answer:
Calculating the excess reactant left after the reaction between sulfur (S) and fluorine (F2) shows that sulfur is in excess by 20.5 grams. The correct answer is option a.
Explanation:
To calculate the mass of excess reactant left when sulfur (S) reacts with fluorine (F2) to form sulfur hexafluoride (SF6), we first need to find the limiting reactant by comparing the mole ratio from the balanced chemical equation and the molar quantities of reactants present.
The balanced chemical reaction is: S(s) + 3F2(g) → SF6(g)
The molar mass of S is 32.07 g/mol, and the molar mass of F2 is approximately 38.00 g/mol. From the provided masses:
- 50.0 g S ÷ 32.07 g/mol = 1.559 mol of S
- 105.0 g F2 ÷ 38.00 g/mol = 2.763 mol of F2
According to the reaction, every mole of S requires 3 moles of F2.
So, 1.559 mol S needs 1.559 mol × 3 = 4.677 mol of F2.
Since we only have 2.763 mol of F2 available, fluorine is the limiting reactant.
The amount of excess S can be calculated assuming all of F2 reacts:
- 2.763 mol F2 / 3 = 0.921 mol of S required
- 0.921 mol S × 32.07 g/mol = 29.5 g of S
Thus, the excess amount of S will be the initial mass minus the mass that reacted:
50.0 g - 29.5 g = 20.5 g S
Hence, the mass of excess reactant leftover is 20.5 g of S, option (a).
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