High School

Sulfur dioxide and oxygen react to form sulfur trioxide, like this:

\[2SO_2 (g) + O_2 (g) \rightarrow 2SO_3 (g)\]

Suppose 220 mmol of \(SO_2\) and 110 mmol of \(O_2\) are added to an empty flask. How much \(SO_3\) will be in the flask at equilibrium?

A. none
B. some, but less than 220 mmol
C. 220 mmol
D. more than 220 mmol

Answer :

Final Answer:

Because the amount of O₂ is limiting, only a fraction of the initial 220 mmol of SO₂ will react, resulting in less than 220 mmol of SO₃ at equilibrium.

Therefore, option b) some, but less than 220 mmol is correct.

Explanation:

According to the given balanced chemical equation, 2 moles of sulfur dioxide (SO₂) react with 1 mole of oxygen (O₂) to form 2 moles of sulfur trioxide (SO₃). Therefore, the stoichiometry of the reaction suggests that the number of moles of SO₃ formed would be determined by the limiting reactant, which in this case is the reactant present in lesser quantity.

Initially, there are 220 mmol of SO₂ and 110 mmol of O₂. Since the balanced equation requires 2 moles of SO₂ to react with 1 mole of O₂, the available O₂ is sufficient to react with 55 mmol (110 mmol / 2) of SO₂.

This means that 55 mmol of SO₂ will react, leaving 165 mmol (220 mmol - 55 mmol) of unreacted SO₂. Consequently, only 55 mmol of SO₃ (110 mmol of SO₂ reacted / 2) will be formed at equilibrium, leaving less than the initial 220 mmol of SO₂.

Therefore, option b) some, but less than 220 mmol is correct.

Learn more about Equilibrium

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