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Hydrogen sulfide gas gives rotten eggs their terrible odor. Hydrogen sulfide burns in air (O2) to produce sulfide dioxide gas and water vapor. In 1.38 moles of sulfide dioxide are produced when this reaction occurs, how many moles of oxygen were required?

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maacastrobr

Answer:

2.07 mol Oâ‚‚

Explanation:

First we need to write down the species present in the chemical equation, using the information given by the exercise:

  • Hâ‚‚S + Oâ‚‚ → SOâ‚‚ + Hâ‚‚O

However this equation is not balanced, so now we balance it:

  • 2Hâ‚‚S + 3Oâ‚‚ → 2SOâ‚‚ + 2Hâ‚‚O

Now we can use the stoichiometric ratio to calculate the moles of oxygen from the moles of sulfide dioxide:

  • 1.38 molSOâ‚‚ * [tex]\frac{3molO_{2}}{2molSO_{2}}[/tex] = 2.07 mol Oâ‚‚
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Based on the stoichiometry of the reaction, 2.07 moles of Oâ‚‚ are required to produce 1.38 mole of SOâ‚‚.

What is the balanced equation of the reaction?

The balanced equation of the reaction of Hydrogen sulfide burning in air (O2) to produce sulfide dioxide gas and water vapor. is given below:

  • 2 Hâ‚‚S + 3 Oâ‚‚ → 2 SOâ‚‚ + 2 Hâ‚‚O

The stoichiometric ratio of the equation is then used to calculate the moles of oxygen from the moles of sulfide dioxide:

2 moles of SOâ‚‚ are produced from 3 moles of Oâ‚‚

1.38 moles of SOâ‚‚ will be produced from x moles of Oâ‚‚

x = 1.38 × 3/2

x = 2.07 moles of Oâ‚‚ will be required

Therefore, 2.07 moles of Oâ‚‚ are required to produce 1.38 mole of SOâ‚‚.

Learn more about stoichiometry at: https://brainly.com/question/16060223

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