SO₂ reacts with H₂S as follows: 2H₂S + SO₂ $\to$ 3S + 2H₂O When 7.50 g of H₂S reacts with 12.75 g of SO₂,which statement applies? A)6.38 g of sulfur are formed. B)10.6 g of sulfur are formed. C)0.0216 moles of H₂S remain. D)1.13 g of H₂S remain. E)SO₂ is the limiting reagent.

Question

Quizplus · Accepted Answer

The balanced reaction shows that 2 moles of H₂S react with 1 mole of SO₂ to produce 3 moles of sulfur. Calculating moles: 7.50 g H₂S (0.220 moles) and 12.75 g SO₂ (0.199 moles). SO₂ is the limiting reagent, producing 0.199 moles of sulfur, which is 10.6 g.

SO2 Reacts with H2S as Follows: 2H2S + SO2 →\to→ 3S + 2H2O When 7

SO₂ Reacts with H₂S as Follows: 2H₂S + SO₂ $\to$ 3S + 2H₂O
When 7