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How many liters of O2 are needed to react completely with 45.0 L of H2S at STP? (2 points) 2H2S (g) + 3O2 (g) 2SO2 (g) + 2H2O (g)

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MissPhiladelphia
The balanced chemical reaction is given as:

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

We are given the volume at STP of H2S to be used for the reaction. This would be the starting point for the calculations. We do as follows:

45.0 L H2S ( 1 mol / 22.4 L) ( 3 mol O2 / 2 mol H2S ) ( 22.4 L / 1 mol ) = 67.5 L O2 needed 
Eduard22sly

The volume of O₂ needed for the reaction is 67.5 L

Data obtained from the question

  • 2H₂S + 3O₂ —> 2SO₂ + 2H₂O
  • Volume of H₂S = 45 L
  • Volume of O₂ =?

From the balanced equation above,

2 L of H₂S required 3 L of O₂

How to determine the volume of O₂

2 L of H₂S required 3 L of O₂.

Therefore,

45 L of H₂S will require = (45 × 3) / 2 = 67.5 L of O₂

Thus, 67.5 L of O₂ is needed for the reaction

Learn more about stoichiometry:

https://brainly.com/question/14735801

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