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Answer :
Final answer:
The mass of excess reactant left after 50.0 g of Sulfur (S) reacts with 105.0 g of Fluorine (F2) to form Sulfur Hexafluoride (SF6) according to the given balanced equation is 20.5 g of Sulfur. This is because Fluorine is the limiting reactant in this reaction, causing some initial Sulfur to remain unreacted.
This correct answer is 1)
Explanation:
To solve this question, we need to find out which of the two reactants is the limiting reactant as it will determine the amount of excess reactant remaining. Sulfur hexafluoride (SF6) is formed from the reaction of sulfur (S) with fluorine (F2) according to the balanced chemical equation, S + 3F2 → SF6.
First, we calculate the number of moles of S and F2. For S, 50g divided by its molar mass (~32g/mol) is 1.56 mol. For F2, 105g divided by its molar mass (38g/mol) is about 2.76 mol.
According to the balanced chemical equation 1 mol S reacts with 3 mol F2. We have more than 1 mol S and less than 3 mol F2, so F2 is the limiting reactant. This means there will be excess S remaining after the reaction.
To find out exactly how much S is left, we need to calculate how much S reacts with the available F2. From the balanced equation, 1 mol S only requires 3 mol F2. Thus, 2.76 mol F2 will react with 2.76/3 mol S, which is 0.92 mol S. Subtracting these 0.92 mol from our initial 1.56 mol S, we get 0.64 mol S unreacted.
Finally, to find this mass in grams, we convert these moles back into grams (0.64 mol * 32g/mol) which gives around 20.5g. From the given options, we thus select 1) 20.5 g S as the mass of excess reactant left after the reaction.
This correct answer is 1)
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