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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.

Respuesta :

Dejanras

The answer is: the percent yield of the reaction is 91.77%.

Balanced chemic reaction: 2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g).

n(H₂S) = 8.3 mol; amount of hydrogen sulfide.

m₁(H₂O) = 137.1 g; mass of water.

From chemical reaction: n(H₂S) : n(H₂O) = 2 : 2 (1 : 1).

n(H₂O) = 8.3 mol; amount of water.

m(H₂O) = 8.3 mol · 18 g/mol.

m(H₂O) = 149.4 g.

the percent yield = 137.1 g ÷ 149,4 g · 100%.

the percent yield = 91.77%.

Eduard22sly

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 of H₂O =?

Mass = mole × molar mass

Mass of H₂O = 8.3 × 18

Mass of H₂O = 149.4 g

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 =?

[tex]Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{137.1}{149.4} * 100\\\\[/tex]

Percentage yield of H₂O = 91.8%

Therefore, the percentage yield of H₂O is 91.8%

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