Consider this balanced chemical equation: 5SF4 + 2I2O5 = 4IF5 + 5SO2 a. What is the limiting reagent when 4.687 grams SF4 reacts with 6.281 grams I2O5 to produce IF5 and SO2.

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.

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