The percentage of in the original mixture was: — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question

💡 Solution & Explanation

$\text{Moles of } O_2 = 0.16 / 32 = 0.005\text{ mol}$. <br>From reaction: $\text{moles of } PbO_2 = 2 \times \text{moles of } O_2 = 0.01\text{ mol}$. <br>$\text{Mass of } PbO_2 = 0.01 \times 239 = 2.39\text{ g}$. <br>$\% = (2.39 / 5.0) \times 100 = 47.8\%$.