Final answer:
To produce 216 grams of sulfur dioxide, you would need to consume approximately 161.92 grams of oxygen gas, calculated through stoichiometric relationships based on the balanced chemical equation.
Step-by-step explanation:
The reaction of hydrogen sulfide gas (H₂S) with oxygen (O₂) to produce sulfur dioxide (SO₂) and water vapor is a combustion reaction. To determine how many grams of oxygen gas would be consumed to produce 216 grams of sulfur dioxide, we can use stoichiometry based on the balanced chemical equation:
2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(g)
From the balanced equation, we see that 2 moles of SO₂ are produced from 3 moles of O₂.
First, we need to calculate the moles of SO₂ produced from 216 grams:
- Molar mass of SO₂ = 32.07 (S) + 2(16.00) (O) = 64.07 g/mol
- Moles of SO₂ = 216 g / 64.07 g/mol ≈ 3.37 moles SO₂
Now, we use the stoichiometric ratios to calculate the moles of oxygen gas needed:
- Moles of O₂ required = (3 moles O₂ / 2 moles SO₂) × 3.37 moles SO₂ ≈ 5.06 moles O₂
Finally, we calculate the grams of O₂ from the moles:
- Molar mass of O₂ = 2(16.00) = 32.00 g/mol
- Grams of O₂ = 5.06 moles O₂ × 32.00 g/mol = 161.92 grams O₂
Therefore, approximately 161.92 grams of oxygen gas will be consumed to produce 216 grams of sulfur dioxide.