Final answer:
After a stoichiometry calculation, it is determined that fluorine is the limiting reactant and sulfur is in excess. The mass of excess sulfur remaining would be 22.10 g if no error exists in the options provided.
Step-by-step explanation:
To determine the mass of excess reactant left after a reaction between sulfur (S) and fluorine (F2), we need to perform a stoichiometry calculation to find out which reactant is the limiting reagent. The chemical equation for the reaction of sulfur with fluorine to produce sulfur hexafluoride (SF6) is:
S(s) + 3F2(g) → SF6(g)
First, we convert the masses of the reactants to moles:
- 51.5 g S / (32.07 g/mol) = 1.60 mol S
- 104.4 g F2 / (38.00 g/mol) = 2.75 mol F2
Next, we compare the mole ratio from the balanced equation to the moles we have:
For every mole of S, we need 3 moles of F2. So for 1.60 mol S, we need 4.80 mol F2 (1.60 mol S × 3 mol F2/mol S). Since we only have 2.75 mol F2, fluorine is the limiting reactant, and sulfur will be in excess.
We can now calculate the mass of sulfur that will react completely with the available fluorine:
2.75 mol F2 × (1 mol S / 3 mol F2) × (32.07 g/mol S) = 29.40 g S
Therefore, we subtract this from the initial mass of sulfur:
51.5 g S - 29.40 g S = 22.10 g S remaining
As the result is not available in the given options, we need to check our calculations or the question data again to ensure there is no mistake. If the provided data is correct, and assuming an option error, 22.10 g would be the mass of S remaining after the reaction with the 104.4 g of F2.