Question

4 moles of SO₂ and 6 moles of O₂ react to form SO3. Total number of moles of reactant and product in the container when only 70% SO2 is used is;

(1) 5.5
(2) 8.6
(3) 10.5 1 2 3 4 1 2 3 4
(4) 6.9 O O​

Answer 1

The total number of moles of reactants and products remaining in the container after 70% of SO₂ reacts with O₂ to form SO₃ is 8.6 moles, including unreacted SO₂ and O₂ as well as the formed SO₃.

Step-by-step explanation:

The question involves calculating the total number of moles of reactant and product in a container after a reaction where only 70% of SO₂ is used to form SO₃, according to the balanced chemical equation:

2SO₂ + O₂ → 2SO₃

Originally, there are 4 moles of SO₂ and 6 moles of O₂. If only 70% of the SO₂ reacts, this means that 70% of 4 moles, which is 2.8 moles of SO₂, will react with O₂ to form SO₃. Using the stoichiometry of the balanced equation, for every 2 moles of SO₂ that react, 2 moles of SO₃ are produced, and 1 mole of O₂ is consumed.

Therefore, (2.8 moles of SO₂ / 2 moles of SO₂) × 2 moles of SO₃ = 2.8 moles of SO₃ produced, and 2.8 moles SO₂ / 2 = 1.4 moles of O₂ consumed. The remaining SO₂ is 4 moles - 2.8 moles = 1.2 moles SO₂, and the remaining O₂ is 6 moles - 1.4 moles = 4.6 moles O₂. Hence, the total number of moles of reactants and products is 1.2 moles SO₂ + 4.6 moles O₂ + 2.8 moles SO₃ = 8.6 moles.