A 20.0 mL solution containing 0.2 g impure H O reacts completely with 0.316 g of KMnO in acid soluti — Hydrogen Chemistry Question

💡 Solution & Explanation

**Step 1: Write the balanced equation** In acidic solution, H₂O₂ is oxidized by KMnO₄: $$2\text{KMnO}_4 + 5\text{H}_2\text{O}_2 + 3\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 5\text{O}_2 + 8\text{H}_2\text{O}$$ From the equation: **2 mol KMnO₄ reacts with 5 mol H₂O₂** **Step 2: Calculate moles of KMnO₄** $$\text{Moles of KMnO}_4 = \frac{0.316}{158} = 0.002 \text{ mol}$$ **Step 3: Calculate moles of pure H₂O₂ that reacted** Using the mole ratio (5:2): $$\text{Moles of H}_2\text{O}_2 = 0.002 \times \frac{5}{2} = 0.005 \text{ mol}$$ **Step 4: Calculate mass of pure H₂O₂** $$\text{Mass of pure H}_2\text{O}_2 = 0.005 \times 34 = 0.17 \text{ g}$$ **Step 5: Calculate purity percentage** $$\text{Purity\%} = \frac{\text{Mass of pure H}_2\text{O}_2}{\text{Mass of impure sample}} \times 100$$ $$\text{Purity\%} = \frac{0.17}{0.2} \times 100 = 85\%$$ Therefore, the answer is 85.00.