1. **State the problem:** Calculate the mass of oxygen gas (O2) needed to produce 3.80 g of potassium chlorate (KClO3) with a percentage yield of 86.0% given the reaction: $$2\text{KCl} + 3\text{O}_2 \rightarrow 2\text{KClO}_3$$ 2. **Write the molar masses:** - Molar mass of KClO3 = $39.10 + 35.45 + 3 \times 16.00 = 122.55$ g/mol - Molar mass of O2 = $2 \times 16.00 = 32.00$ g/mol 3. **Calculate moles of KClO3 produced (actual):** $$\text{moles KClO}_3 = \frac{3.80}{122.55} = 0.0310 \text{ mol}$$ 4. **Calculate moles of KClO3 expected (theoretical) using percentage yield:** $$\text{moles KClO}_3^{\text{theoretical}} = \frac{0.0310}{0.86} = 0.0360 \text{ mol}$$ 5. **Use stoichiometry to find moles of O2 needed:** From the balanced equation, $3$ moles of O2 produce $2$ moles of KClO3. $$\text{moles O}_2 = 0.0360 \times \frac{3}{2} = 0.0540 \text{ mol}$$ 6. **Calculate mass of O2 needed:** $$\text{mass O}_2 = 0.0540 \times 32.00 = 1.73 \text{ g}$$ **Final answer:** The mass of oxygen gas needed is **1.73 g**.