Final answer:
The mass of the excess reactant left over is -7.12 g. This means that there is an excess of F2 and no excess of S left over.
Step-by-step explanation:
To determine the mass of the excess reactant left over, we first need to identify the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and limits the amount of product that can be formed. We can calculate the moles of each reactant by dividing the given mass by the molar mass.
For S: 51.7 g / (16.00 g/mol) = 3.23 mol
For F2: 103.7 g / (38.00 g/mol) = 2.73 mol
From the balanced equation:
2S + 3F2 → 2SF6
Since the stoichiometric ratio is 2:3, we can see that F2 is the limiting reactant. Therefore, all of the F2 will be consumed, and we need to calculate the mass of the excess reactant (S) that is not consumed:
Excess mass of S = (moles of S) x (molar mass of S) - (moles of F2) x (molar mass of S) = (3.23 mol) x (32.00 g/mol) - (2.73 mol) x (32.00 g/mol) = 80.24 g - 87.36 g = -7.12 g
Since the result is negative, it means that there is an excess of F2 and no excess of S left over.