Question

Hydrogen sulfide gas burns in oxygen to produce sulfur dioxide and water vapor:

2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(g)

What mol of oxygen gas is consumed in a reaction that produces 3 mol of SO₂?

a) 3 mol
b) 4.5 mol
c) 1.5 mol
d) 6 mol

Answer 1

In the balanced chemical equation, 2 moles of hydrogen sulfide react with 3 moles of oxygen to produce 2 moles of sulfur dioxide. The ratio of moles of oxygen to moles of sulfur dioxide is 3:2. To determine the moles of oxygen gas consumed in a reaction that produces 3 moles of sulfur dioxide, we can set up a proportion and solve for x. The correct answer is b) 4.5 mol.

Step-by-step explanation:

The balanced chemical equation for the reaction is:

2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(g)

From the balanced equation, we can see that for every 2 moles of hydrogen sulfide (H₂S) that react, 3 moles of oxygen (O₂) are consumed to produce 2 moles of sulfur dioxide (SO₂). This means that the ratio of moles of oxygen to moles of sulfur dioxide is 3:2.

To determine the moles of oxygen gas consumed in a reaction that produces 3 moles of sulfur dioxide, we can set up the following proportion:

(x mol O₂) / (3 mol SO₂) = (3 mol O₂) / (2 mol SO₂)

Cross-multiplying, we have:

2x = 9

x = 4.5 mol

Therefore, the correct answer is b) 4.5 mol.