Which of the following is the correct "setup" for the problem "How many grams of H₂O will be produced from 2.1 moles of O₂ and an excess of H₂S?" according to the reaction 2H₂S + 3O₂ 2H₂O + 2SO₂

Question

Quizplus · Accepted Answer

The correct setup for the problem is as follows: - 2.1 moles of O₂ - Excess of H₂S Use the mole ratio from the balanced chemical equation to convert the given amount of O₂ to H₂S: 3 moles of O₂ : 2 moles of H₂S Thus, the amount of H₂S in excess can be calculated as: 2.1 moles of O₂ x (2 moles of H₂S/3 moles of O₂) = 1.4 moles of H₂S Now, use the mole ratio from the balanced chemical equation to calculate the moles of H₂ produced: 2 moles of H₂S : 2 moles of SO₂ : 1 mole of H₂O Since there is excess H₂S, SO₂ is the limiting reactant. So, the number of moles of H₂ produced is: 2.1 moles of O₂ x (2 moles of H₂S/3 moles of O₂) x (2 moles of SO₂/2 moles of H₂S) x (1 mole of H₂O/2 moles of SO₂) = 0.7 moles of H₂O Finally, use the molar mass of H₂O to convert moles to grams: 0.7 moles of H₂O x 314.43 g/mol = 220.1 g of H₂O The correct choice is C.

Which of the Following Is the Correct "Setup" for the Problem