Question

If O2 was the excess reagent, 8.3 mol of H2S were consumed, and 13.7 moles of water were collected after the reaction has gone to completion, what is the theoretical yield of Sulfate in moles? ROUND TO THE NEAREST TENTH!!

a. 26.7 moles
b. 37.0 moles
c. 28.6 moles
d. 23.3 moles

Answer 1

The theoretical yield of sulfate cannot be determined without the precise chemical reaction for the formation of sulfate from hydrogen sulfide (H2S) and oxygen (O2), as the stoichiometry for the production of sulfate is not provided in the question.

Step-by-step explanation:

To calculate the theoretical yield of sulfate, we need to understand the stoichiometry of the reaction in question. Let's consider a similar reaction of hydrogen sulfide and oxygen:

2H₂S (g) + 3O₂ (g) → 2SO₂ (g) + 2H₂O(g)

We know that 8.3 moles of H₂S were consumed and 13.7 moles of H₂O were produced. From the balanced equation, the mole ratio of H₂S to SO₂ is 1:1. Therefore, 8.3 moles of H₂S would produce 8.3 moles of SO₂ theoretically. However, the reaction for producing sulfate (SO₄²-) is different, and it would involve further oxidation of SO₂, which is not specified in the question. If the reaction simply produced SO₄²-, the mole ratio of H₂S to SO₄²- would be different. Without the exact reaction provided for the formation of sulfate, the theoretical yield of sulfate cannot be directly determined from the provided information.