2 KClO3(s) → 2 KCl(s) + 3 O2(g) What is the percentage yield of O2 if 12.3 g of KClO3 (molar mass 12 — Stoichiometry Chemistry Question

STEPS:

1. Calculate the Moles of Reactant: Begin by determining how many moles of potassium chlorate ($KClO_3$) were used. Divide the given mass by its molar mass:
$$\text{Moles of } KClO_3 = \frac{12.3 \text{ g}}{123 \text{ g/mol}} = \mathbf{0.100 \text{ mol } KClO_3}$$
2. Determine the Theoretical Yield (in moles): Use the stoichiometric coefficients from the balanced equation ($2 KClO_3 \rightarrow 2 KCl + 3 O_2$) to find how many moles of oxygen gas ($O_2$) *should* have been produced. The ratio of $KClO_3$ to $O_2$ is $2:3$.
$$\text{Theoretical moles of } O_2 = 0.100 \text{ mol } KClO_3 \times \left(\frac{3 \text{ mol } O_2}{2 \text{ mol } KClO_3}\right) = \mathbf{0.150 \text{ mol } O_2}$$
3. Calculate the Actual Yield (in moles): Convert the actual mass of $O_2$ produced into moles using its molar mass:
$$\text{Actual moles of } O_2 = \frac{3.2 \text{ g}}{32 \text{ g/mol}} = \mathbf{0.10 \text{ mol } O_2}$$
4. Calculate the Percentage Yield: The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100%:
$$\text{Percentage Yield} = \left(\frac{\text{Actual moles}}{\text{Theoretical moles}}\right) \times 100\% = \left(\frac{0.10 \text{ mol}}{0.15 \text{ mol}}\right) \times 100\%$$
$$\text{Percentage Yield} = \frac{2}{3} \times 100\% \approx \mathbf{67\%}$$
5. Conclusion: The calculated yield of 67% matches option B.

WHY_OTHERS_WRONG:

• A) 100%: This would assume that the actual mass of $O_2$ produced was exactly equal to the theoretical mass ($4.8 \text{ g}$). Since the student only collected $3.2 \text{ g}$, the yield must be less than 100%.
• C) 50%: This might be chosen if a student incorrectly used a $1:1$ stoichiometric ratio ($0.10 \text{ mol } O_2$ actual vs. $0.20 \text{ theoretical}$) or made a significant computational error.
• D) 33%: This value is the complement of the correct answer ($100\% - 67\% = 33\%$), representing the percentage of product *lost* rather than the percentage yielded.
• E) 10%: This likely results from a student simply looking at the $0.10 \text{ mol}$ of $KClO_3$ calculated in Step 1 and confusing the numerical value with a percentage.