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

Quizplus · Accepted Answer

To determine the limiting reactant, we need to calculate the number of moles of each reactant: moles of H₂S = 7.50 g / (32 g/mol + 2(32 g/mol) + 32 g/mol + 32 g/mol + 32 g/mol) = 0.0216 mol moles of SO₂ = 12.75 g / (32 g/mol + 2(32 g/mol) + 32 g/mol + 32 g/mol + 32 g/mol) = 0.0366 mol We see that there are fewer moles of H₂S, so it is the limiting reactant. Using the balanced equation, we can calculate the amount of sulfur produced: 2 mol H₂S produces 3 mol S 0.0216 mol H₂S produces (3/2)(0.0216) mol S = 0.0324 mol S We can convert mol S to grams: 0.0324 mol S x 32.07 g/mol = 1.04 g S Therefore, the answer is D, 10.6 g of sulfur is formed.

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

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