Consider the unbalanced equation: H2S(s) + O2(g)  SO2(g) + H2O(l)

Question

a. If 178.0 g of H2S are mixed with 157.0 g O2 which reactant is limiting and what is the theoretical yield of sulfur dioxide (in grams)?

b. How much of the excess reactant (in grams) remains?

c. If only 198.7 g of sulfur dioxide were produced in the lab from part a, what was the percent yield?

Answer 1

First, the balanced reaction is 2 H₂S + 3 O₂ --> 2 SO₂ + 2 H₂O.

a) (178.0g H₂S) * (1 mol H₂S / 34.08g H₂S) * (2 mol SO₂ / 2 mol H₂S) = 5.223 mol SO₂.

(157.0g O₂) * (1 mol O₂ / 32.00g O₂) * (2 mol SO₂ / 3 mol O₂) = 3.271 mol SO₂.

Thus oxygen is the limiting reactant, and the theoretical yield is (3.271 mol SO₂) * (64.06g SO₂ / 1 mol SO₂) = 209.5g SO₂.

b) The difference in moles of SO₂ produced is 5.223 - 3.271 = 1.952 mol SO₂, and the mole ratio in the reaction is 1:1 with H₂S, so (1.952 mol H₂S) * (34.08g H₂S / 1 mol H₂S) = 66.5g H₂S remains.

c) Percent yield 100 * actual/theoretical = 100 * (198.7/209.5) = 94.83%.