of a sample of and is heated. The weight of the residue is . What is the percentage of in the mixtur — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question

💡 Solution & Explanation

Let $x$ be mass of $CaCO_3$. $\text{Loss} = 10 - 6.15 = 3.85 \text{ g } (CO_2)$. <br>$\text{moles } CO_2 = \frac{x}{100} + \frac{10-x}{197} = \frac{3.85}{44} = 0.0875$. <br>Solving for $x$: $1.97x + 1000 - 100x = 1723.75 \Rightarrow 0.97x \approx 4.85 \Rightarrow x \approx 5 \text{ g}$. <br>$\% = (5/10) \times 100 = 50\%$.