SO₂ reacts with H₂S as follows:
2H₂S + SO₂ → 3S + 2H₂O
When 7.50 g of H₂S reacts with 12.75 g of SO₂, which statement applies?

Question

SO₂ reacts with H₂S as follows: 2H₂S + SO₂ → 3S + 2H₂O When 7.50 g of H₂S reacts with 12.75 g of SO₂, which statement applies?

Quizplus · Accepted Answer

First, we need to determine which reactant is the limiting reagent. We can do this by calculating how much sulfur can be produced from each reactant separately and comparing those amounts. From H₂S: 7.50 g H₂S x (1 mol / 208.33 g) x (3 mol S / 2 mol H₂S) x (32.06 g / 1 mol) = 3.22 g S From SO₂: 12.75 g SO₂ x (1 mol / 514.54 g) x (3 mol S / 1 mol SO₂) x (32.06 g / 1 mol) = 14.2 g S Since SO₂ can produce more sulfur than H₂S, SO₂ is the limiting reagent. Using the balanced equation, we can determine how much sulfur is produced from the reaction: 2H₂S + SO₂ → 3S + 2H₂O 12.75 g SO₂ x (3 mol S / 1 mol SO₂) x (32.06 g / 1 mol) = 122.6 g S Therefore, 122.6 g - 14.2 g = 108.4 g of sulfur is formed. Converting to grams: 108.4 g S x (1 mol / 32.06 g) x (208.33 g / 1 mol H₂S) = 718.3 g H₂S reacted Subtracting the amount of H₂S used from the original amount gives the remaining amount: 7500 mg - 718.3 g = 31.7 g Converting to grams: 31.7 g x (1 mol / 208.33 g) x (514.54 g / 1 mol SO₂) = 2.44 g SO₂ remaining Therefore, the correct answer is D) 10.6 g of sulfur is formed.

SO2 Reacts with H2S as Follows

SO₂ Reacts with H₂S as Follows