To determine the amount in moles of [tex]\(\text{SO}_3\)[/tex] formed when [tex]\(0.36\)[/tex] moles of [tex]\(\text{SO}_2\)[/tex] react with excess [tex]\(\text{O}_2\)[/tex], we follow these steps:
1. Understand the balanced chemical equation:
[tex]\[
2 \text{SO}_2 + \text{O}_2 \rightarrow 2 \text{SO}_3
\][/tex]
2. Identify the stoichiometric relationship:
From the balanced equation, we see that 2 moles of [tex]\(\text{SO}_2\)[/tex] produce 2 moles of [tex]\(\text{SO}_3\)[/tex]. This means there is a 1:1 molar ratio between [tex]\(\text{SO}_2\)[/tex] and [tex]\(\text{SO}_3\)[/tex].
3. Apply the stoichiometric relationship:
Given that there is a 1:1 ratio, the moles of [tex]\(\text{SO}_3\)[/tex] produced will be equal to the moles of [tex]\(\text{SO}_2\)[/tex] reacted, assuming all of the [tex]\(\text{SO}_2\)[/tex] is completely converted into [tex]\(\text{SO}_3\)[/tex].
4. Calculate the moles of [tex]\(\text{SO}_3\)[/tex] formed:
Since we start with [tex]\(0.36\)[/tex] moles of [tex]\(\text{SO}_2\)[/tex], and each mole of [tex]\(\text{SO}_2\)[/tex] produces an equivalent mole of [tex]\(\text{SO}_3\)[/tex], the moles of [tex]\(\text{SO}_3\)[/tex] formed will also be [tex]\(0.36\)[/tex] moles.
Therefore, the amount in moles of [tex]\(\text{SO}_3\)[/tex] formed is:
[tex]\[
0.36 \text{ moles}
\][/tex]
