Sulfur and oxygen react to produce sulfur trioxide.In a particular experiment,7.9 grams of SO₃are produced by the reaction of 6.0 grams of O₂ with 7.0 grams of S.What is the % yield of SO₃ in this experiment?
S (s) + O₂(g) → SO₃ (g) (not balanced)

Question

Quizplus · Accepted Answer

First, we need to balance the equation to determine the mole ratio between S and SO2:
2S (s) + 3O2 (g) → 2SO3 (g)

Next, we need to calculate the theoretical yield of SO3 using the mole ratio and the limiting reactant (S):
Moles of S used = 7.0 g / 32.06 g/mol = 0.218 mol
Moles of SO3 produced = 2 * 0.218 mol = 0.436 mol
Theoretical yield of SO3 = 0.436 mol * 80.06 g/mol = 34.92 g

Finally, we can calculate the percent yield:
% Yield = (Actual yield / Theoretical yield) x 100%
Actual yield = 7.9 g
% Yield = (7.9 g / 34.92 g) x 100% = 22.6%

The closest answer choice to this calculation is D) 79.

Sulfur and Oxygen React to Produce Sulfur Trioxide