of a sample of was strongly heated and the weight of the residue was . The percentage purity of in t — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question

💡 Solution & Explanation

$CaCO_3 \rightarrow CaO + CO_2$. Loss in weight $= \text{mass of } CO_2 = 1 - 0.78 = 0.22\text{ g}$. <br>$\text{Moles of } CO_2 = 0.22 / 44 = 0.005$. $\text{Moles of } CaCO_3 \text{ reacted} = 0.005$. <br>$\text{Mass of } CaCO_3 = 0.005 \times 100 = 0.5\text{ g}$. $\% \text{ purity} = (0.5/1) \times 100 = 50\%$.