Let's solve the problem step-by-step to find the percent yield of oxygen in this reaction.
1. Understand the reaction:
The balanced chemical equation is:
[tex]\[
2 \text{KClO}_3 \rightarrow 2 \text{KCl} + 3 \text{O}_2
\][/tex]
2. Given data:
- Actual yield of [tex]\(\text{O}_2\)[/tex]: [tex]\(115.0\)[/tex] grams
- Initial amount of potassium chlorate ([tex]\(\text{KClO}_3\)[/tex]): [tex]\(400.0\)[/tex] grams
- Molar mass of [tex]\(\text{KClO}_3\)[/tex]: [tex]\(122.55\)[/tex] g/mol
- Molar mass of [tex]\(\text{O}_2\)[/tex]: [tex]\(32.00\)[/tex] g/mol
3. Calculate moles of [tex]\(\text{KClO}_3\)[/tex]:
[tex]\[
\text{Moles of } \text{KClO}_3 = \frac{\text{Mass of KClO}_3}{\text{Molar mass of KClO}_3} = \frac{400.0 \text{ g}}{122.55 \text{ g/mol}} \approx 3.26397 \text{ mol}
\][/tex]
4. Relate moles of [tex]\(\text{KClO}_3\)[/tex] to moles of [tex]\(\text{O}_2\)[/tex]:
According to the balanced equation, 2 moles of [tex]\(\text{KClO}_3\)[/tex] produce 3 moles of [tex]\(\text{O}_2\)[/tex], thus:
[tex]\[
\text{Moles of } \text{O}_2 = \left( 3.26397 \text{ mol} \times \frac{3 \text{ mol O}_2}{2 \text{ mol KClO}_3}\right) \approx 4.89596 \text{ mol}
\][/tex]
5. Calculate theoretical yield of [tex]\(\text{O}_2\)[/tex] in grams:
[tex]\[
\text{Theoretical yield of } \text{O}_2 = \text{Moles of } \text{O}_2 \times \text{Molar mass of } \text{O}_2 = 4.89596 \text{ mol} \times 32.00 \text{ g/mol} \approx 156.67075 \text{ g}
\][/tex]
6. Calculate the percent yield:
[tex]\[
\text{Percent yield} = \left(\frac{\text{Actual yield}}{\text{Theoretical yield}}\right) \times 100 = \left(\frac{115.0 \text{ g}}{156.67075 \text{ g}}\right) \times 100 \approx 73.40 \%
\][/tex]
Therefore, the percent yield of oxygen in this chemical reaction is [tex]\(73.40\%\)[/tex]. The correct answer is [tex]\(\boxed{73.40\%}\)[/tex].
