Final answer:
The total number of moles of reactant and product in the container, after 70% of the SO₂ reacts with O₂ to form SO₃, is 10.2 moles, not 6.9 as stated in the question.
Step-by-step explanation:
The student is asking about the stoichiometry related to the reaction between SO₂ and O₂ to form SO₃. Given that only 70% of SO₂ is used, we must first calculate the amount of SO₂ reacting. Starting with 4 moles of SO₂, 70% use translates to 2.8 moles of SO₂ reacting. The balanced equation is 2SO₂ + O₂ →; 2SO₃, meaning for every 2 moles of SO₂ used, 1 mole of O₂ is needed.
Therefore, 2.8 moles of SO₂ would require 1.4 moles of O₂. From the original 6 moles of O₂, we subtract the 1.4 moles that reacted, leaving us with 4.6 moles of unreacted O₂. The product formed is also 2.8 moles of SO₃ since the mole ratio of SO₂ to SO₃ is 1:1. In total, the number of moles of reactants and products is 2.8 + 4.6 + 2.8, which equals 10.2 moles. However, the error in the student's answer seems to be that they considered only the reactant SO₂ and product SO₃ but did not properly account for the unused O₂. Correcting the error, the total number of moles should be 10.2, not 6.9.