Answer:
The limiting reagent is the SF₄
Explanation:
In order to determine the limiting reagent we convert the mass of the reactants to moles, and then, we work with stoichiometry.
4.687 g / 108.06 g/mol = 0.0433 moles of SF₄
6.281 g / 333.8 g/mol = 0.0188 moles of I₂O₅
The reaction is: 5SF₄ + 2I₂O₅ = 4IF₅ + 5SO₂
Ratio is 5:2. 5 moles of fluoride react with 2 moles of pentoxide
Then, 0.0433 moles of fluoride will react with (0.0433 . 2) / 5 = 0.0173 moles
We have 0.0188 moles of I₂O₅ and we need 0.0173 so there are some moles of pentoxide that remains after the reaction. In conclussion, the limiting reagent is the SF₄. We verify:
2 moles of pentoxide react with 5 moles of SF₄
Therefore, 0.0188 moles of I₂O₅ will react with (0.0188 . 5) / 2 = 0.0470 moles.
As we have 0.0433 moles of SF₄, we do not have enough moles.
