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Answer :
The correct answer is the letter "a", 20.5 g S. The excess reactant is sulfur, so the mass that exceeds the reaction given is 20.51 grams of sulfur.
Limiting and excess reactant
The balanced chemical equation for the reaction between sulfur and fluorine to form sulfur hexafluoride is:
- S(s) + 3F₂(g) → SF₆(g)
To solve this problem, we need to first determine which reactant is limiting and which one is in excess. The limiting reactant is the one that is completely used up in the reaction, while the excess reactant is the one that is left over.
To find the limiting reactant, we need to use stoichiometry to calculate the amount of product that can be formed from each reactant. We can do this by using the molar mass of each reactant and the balanced chemical equation.
- The molar mass of sulfur (S) is 32.06 g/mol. Therefore, 50.0 g of S is equivalent to 50.0 g / 32.06 g/mol = 1.56 mol of S.
- The molar mass of fluorine (F₂) is 38.00 g/mol. Therefore, 105.0 g of F₂) is equivalent to 105.0 g / 38.00 g/mol = 2.76 mol of F₂.
Using the balanced chemical equation, we can see that 1 mol of S reacts with 3 mol of F₂ to produce 1 mol of SF₆. Therefore, the maximum amount of SF₆ that can be produced from the given amounts of S and F₂ is:
- From 1.56 mol of S, we can produce 1.56/1 x 1 = 1.56 mol of SF₆
- From 2.76 mol of F₂, we can produce 2.76/3 x 1 = 0.92 mol of SF₆
Since the reaction requires 3 moles of F₂ for every mole of S, we can see that there is not enough F₂ to react with all of the S. Therefore, F₂ is the limiting reactant and S is the excess reactant.
To calculate the mass of excess reactant left over, we need to first determine how much of the excess reactant reacted. Since all of the F₂ reacted, we can use the balanced chemical equation to calculate how much S reacted:
- 3 mol of F₂ reacts with 1 mol of S
- 2.76 mol of F₂ reacted, so the amount of S that reacted is (2.76 mol F₂) / (3 molF₂/mol S) = 0.92 mol of S
The amount of excess S left over is therefore:
- 1.56 mol of S (initial amount) - 0.92 mol of S (the amount that reacted) = 0.64mol of S
Finally, we can calculate the mass of excess S left over:
- Mass of excess S = (0.64 mol S) x (32.06 g/mol S) = 40.08 g of S
Therefore, 20.51 g of S is the mass of excess reactant left over and the correct answer is letter A) 20.5 g S
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