Final answer:
To determine how many grams of sulfur dioxide will be produced, we perform stoichiometric calculations to find that 29.1 grams of sulfur dioxide would be produced from the reaction of 23.9 g of hydrogen sulfide with 21.8 g of oxygen.
Step-by-step explanation:
The student asked how many grams of sulfur dioxide would be produced if 23.9 g of hydrogen sulfide gas reacted with 21.8 g of oxygen gas. First, we need the balanced chemical equation for the reaction of hydrogen sulfide with oxygen, which is:
2H₂S (g) + 3O₂ (g) → 2SO₂ (g) + 2H₂O(g)
To solve this problem, we will use stoichiometry to convert the mass of reactants to moles, use the mole ratio from the balanced equation, and then convert the moles of sulfur dioxide produced back to grams.
Step 1: Convert the mass of reactants to moles
The molar mass of H₂S is approximately 34.08 g/mol and the molar mass of O₂ is approximately 32.00 g/mol.
23.9 g H₂S * (1 mol H₂S / 34.08 g H₂S) = 0.701 mol H₂S
21.8 g O₂ * (1 mol O₂ / 32.00 g O₂) = 0.681 mol O₂
Step 2: Determine the limiting reactant
Using the mole ratio from the balanced equation, 2 moles of H₂S react with 3 moles of O₂. Therefore, for 0.701 moles of H₂S, we would need 1.0515 moles of O₂. But we only have 0.681 moles of O₂, which makes oxygen the limiting reactant.
Step 3: Calculate the moles of SO₂ produced
According to the stoichiometry of the reaction, 3 moles of O₂ produce 2 moles of SO₂.
0.681 mol O₂ * (2 mol SO₂ / 3 mol O₂) = 0.454 mol SO₂
Step 4: Convert moles of SO₂ to grams
The molar mass of SO₂ is approximately 64.07 g/mol.
0.454 mol SO₂ * (64.07 g SO₂ / 1 mol SO₂) = 29.1 g SO₂
Thus, 29.1 grams of sulfur dioxide would be produced from the reaction of 23.9 g of hydrogen sulfide with 21.8 g of oxygen.