Question

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?

Answer 1

Answer:

2.07 mol O₂

Explanation:

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

• H₂S + O₂ → SO₂ + H₂O

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

• 2H₂S + 3O₂ → 2SO₂ + 2H₂O

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

• 1.38 molSO₂ * $\frac{3molO_{2}}{2molSO_{2}}$ = 2.07 mol O₂

Answer 2

Based on the stoichiometry of the reaction, 2.07 moles of O₂ are required to produce 1.38 mole of SO₂.

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 H₂S + 3 O₂ → 2 SO₂ + 2 H₂O

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

2 moles of SO₂ are produced from 3 moles of O₂

1.38 moles of SO₂ will be produced from x moles of O₂

x = 1.38 × 3/2

x = 2.07 moles of O₂ will be required

Therefore, 2.07 moles of O₂ are required to produce 1.38 mole of SO₂.

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