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? A)6.38 g of sulfur is formed. B)SO₂ is the limiting reagent. C)0.0216 mol of H₂S remains. D)10.6 g of sulfur is formed. E)1.13 g of H₂S remains.

Question

Quizplus · Accepted Answer

To determine the limiting reagent, we need to convert the given masses of reactants to moles of each. Moles of H₂S = 7.50 g / (32.07 g/mol + 2(32.06 g/mol) + 32.07 g/mol + 32.06 g/mol) = 0.0066 mol Moles of SO₂ = 12.75 g / (32.07 g/mol + 2(32.06 g/mol) + 32.07 g/mol + 32.06 g/mol) = 0.0111 mol Based on the balanced chemical equation, we can see that 2 moles of H₂S react with 1 mole of SO₂. Therefore, if we have more than twice as many moles of H₂S as we do SO₂, the excess H₂S will not react. (2 x 0.0111 mol)/0.0066 mol = 3.54 Since 3.54 is greater than 2, we know that we have excess H₂S and that SO₂ is the limiting reagent. To calculate the amount of sulfur produced, we first need to figure out how many moles of SO₂ will react based on the amount of H₂S available. (0.0066 mol H₂S) / (2 mol H₂S) x (1 mol SO₂ / 2 mol H₂S) = 0.0033 mol SO₂ Now we can calculate the mass of sulfur produced: (0.0033 mol SO₂) x (3 mol S / 2 mol SO₂) x (32.07 g/mol) = 16.08 g sulfur Therefore, the correct answer is D) 10.6 g of sulfur is formed. (Note that choice A is close to the correct answer, but it appears to have been calculated using the mass of H₂S as the limiting reagent, which is not correct in this case.)

SO2 Reacts with H2S as Follows

SO₂ Reacts with H₂S as Follows