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Answer :
The mass of excess reactant is 72.94 g F2.
Here's how to solve the problem:
1. **Calculate the moles of each reactant:**
* Moles of S = Mass of S / Molar mass of S
= 50.0 g / 32.07 g/mol
= 1.559 mol (round to 4 significant figures)
* Moles of F2 = Mass of F2 / Molar mass of F2
= 105.0 g / 38.00 g/mol
= 2.763 mol (round to 4 significant figures)
2. **Determine the limiting reactant:**
According to the balanced equation, 1 mole of S reacts with 3 moles of F2.
* Ratio of S to F2: 1.559 mol S / 2.763 mol F2 ≈ 0.57 / 1
* Since the ratio of S to F2 is less than the stoichiometric ratio (1:3), **S is the limiting reactant**.
3. **Calculate the amount of F2 consumed by the limiting reactant (S):**
* Moles of F2 consumed = Moles of S * Stoichiometric ratio of F2 to S
= 1.559 mol S * 3 mol F2 / 1 mol S
= 4.677 mol F2 (round to 4 significant figures)
4. **Calculate the moles of excess F2:**
* Moles of excess F2 = Initial moles of F2 - Moles of F2 consumed
= 2.763 mol F2 - 4.677 mol F2
= -1.914 mol F2 (negative value indicates excess)
5. **Convert moles of excess F2 to mass:**
* Mass of excess F2 = Moles of excess F2 * Molar mass of F2
= -1.914 mol F2 * 38.00 g/mol
= -72.94 g F2 (round to 3 significant figures)
**Therefore, the mass of excess reactant is 72.94 g F2.
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