Final answer:
The percent yield of sulfur when 22g of H₂S produces 4g of sulfur is not exactly listed in the provided options. By calculating, the theoretical yield is found to be 20.68g of sulfur, and so the percent yield is approximately 19.36%.
Step-by-step explanation:
In the reaction where 22g of hydrogen sulfide (H₂S) is reacted with excess oxygen (O₂) to produce sulfur (S), the student asks what the percent yield of sulfur is when 4g of sulfur is produced. The balanced equation for this reaction is:
2H₂S (g) + 3O₂ (g) → 2SO₂ (g) + 2H₂O(g)
To solve this, we first need to determine the theoretical yield of sulfur that can be produced from 22g of hydrogen sulfide. According to the balanced equation, 2 moles of H₂S produce 2 moles of sulfur, meaning they react on a 1:1 mole basis.
Molar mass of H₂S = 2(1.01) + 32.07 ≈ 34.09 g/mol
Using molar mass to convert from grams to moles:
Moles of H₂S = 22g H₂S ÷ 34.09 g/mol ≈ 0.645 moles of H₂S
Since the mole ratio of H₂S to S is 1:1:
Moles of S = 0.645 moles of H₂S
Using molar mass to convert from moles to grams for sulfur:
Molar mass of S = 32.07 g/mol
Mass of S = 0.645 moles × 32.07 g/mol ≈ 20.68g
This is the theoretical yield of sulfur. Now we can calculate the percent yield:
Percent yield = (Actual yield ÷ Theoretical yield) × 100%
Percent yield = (4g ÷ 20.68g) × 100% ≈ 19.36% (rounded to two decimal places)
This option is not available in the choices provided by the student; they may need to verify their calculations or check if the question was accurately represented.