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44. How much excess reactant remains when 4.687g of sulfur tetrafluoride reacts with 6.281g of diiodine pentoxidernto produce iodine pentafluoride and sulfur dioxide?
Accepted Answer
To determine the excess reactant, we need to perform stoichiometric calculations. Here's a breakdown of the steps:
1. Write the balanced chemical equation:
SF₄ + 5I₂O₅ → 10IF₅ + SO₂
2. Convert the masses of reactants to moles:
Moles of SF₄ = (4.687 g) / (108.06 g/mol) = 0.0434 mol
Moles of I₂O₅ = (6.281 g) / (333.81 g/mol) = 0.0188 mol
3. Determine the limiting reactant:
From the balanced equation, 1 mole of SF₄ reacts with 5 moles of I₂O₅.
The mole ratio of SF₄ to I₂O₅ is 1:5.
The actual ratio of moles of SF₄ to I₂O₅ in the reaction is 0.0434 mol : 0.0188 mol, which is approximately 2.3:1.
Since the actual ratio is greater than the stoichiometric ratio, SF₄ is in excess, and I₂O₅ is the limiting reactant.
4. Calculate the amount of SF₄ consumed by the limiting reactant:
From the balanced equation, 5 moles of I₂O₅ react with 1 mole of SF₄.
Moles of SF₄ consumed = (0.0188 mol I₂O₅)
(1 mol SF₄ / 5 mol I₂O₅) = 0.00376 mol SF₄
5. Calculate the amount of SF₄ remaining:
Moles of SF₄ remaining = (0.0434 mol SF₄) - (0.00376 mol SF₄) = 0.0396 mol SF₄
6. Convert moles of SF₄ remaining to grams:
Mass of SF₄ remaining = (0.0396 mol SF₄)
(108.06 g/mol) = 4.28 g
Therefore, 4.28 g of sulfur tetrafluoride remain as excess reactant.
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