Sulfur dioxide reacts with [tex]\( O_2 \)[/tex] gas to form sulfur trioxide.

a) If 3.4 moles of sulfur dioxide react with excess [tex]\( O_2 \)[/tex] gas, how many moles of sulfur trioxide will form?



Answer :

Sure, let's look at the chemical reaction first. The balanced equation for the reaction between sulfur dioxide (SO₂) and oxygen (O₂) to form sulfur trioxide (SO₃) is:

[tex]\[ 2 \text{SO}_2 + \text{O}_2 \rightarrow 2 \text{SO}_3 \][/tex]

From this balanced equation, we can see that 2 moles of sulfur dioxide (SO₂) react with 1 mole of oxygen (O₂) to form 2 moles of sulfur trioxide (SO₃).

Here's a step-by-step solution to the problem:

1. Identify the moles of SO₂ given:
You have 3.4 moles of sulfur dioxide (SO₂).

2. Relate the moles of SO₂ to SO₃ using the stoichiometric coefficients from the balanced equation:
The balanced equation shows a 1:1 molar ratio between SO₂ and SO₃ (since 2 moles of SO₂ produce 2 moles of SO₃).

3. Calculate the moles of SO₃ formed:
Given that the molar ratio between SO₂ and SO₃ is 1:1, the number of moles of SO₃ formed will be the same as the number of moles of SO₂ reacted.

Therefore, if 3.4 moles of sulfur dioxide react with excess oxygen gas, 3.4 moles of sulfur trioxide (SO₃) will be formed.

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