Consider the following reaction: 2h2s (g) + 3o2 (g)  2so2 (g) + 2h2o (g) if o2 was the excess reagent, 8.3 mol of h2s were consumed, and 137.1 g of water were collected after the reaction has gone to completion, what is the percent yield of the reaction? show your work.

The percentage yield of H₂O when 8.3 moles of H₂S reacts with excess O₂ is 91.8%

We'll begin by calculating the number of mole of H₂O produced by the reaction of 8.3 mole of H₂S. This can be obtained as follow:

2H₂S (g) + 3O₂ (g) —> 2SO₂ (g) + 2H₂O (g)

From the balanced equation above,

2 moles of H₂S reacted to produce 2 moles of H₂S.

Therefore,

8.3 mole of H₂S will also react to produce 8.3 mole of H₂O.

Next, we shall determine the theoretical yield of H₂O by calculating the mass of 8.3 mole of H₂O.

Mole of H₂O = 8.3 moles

Molar mass of H₂O = (1×2) + 16 = 18 g/mol

Mass = mole × molar mass

Mass of H₂O = 8.3 × 18

Thus, the theoretical yield of H₂O is 149.4 g

Finally, we shall determine the percentage yield. This can be obtained as follow:

Actual yield of H₂O = 137.1 g

Theoretical yield of H₂O = 149.4 g

Percentage yield of H₂O = 91.8%

Therefore, the percentage yield of H₂O is 91.8%

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