A sample of and was heated. The loss in weight was . The percentage of in the mixture is: — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question

💡 Solution & Explanation

$2KHCO_3 \rightarrow K_2CO_3 + H_2O + CO_2$. <br>$\text{Loss} = x \times (18+44)/(2 \times 100) = 0.31x$. <br>$0.031 = 0.31x \Rightarrow x = 0.1 \text{ g } KHCO_3$. <br>$K_2CO_3 = 0.4 \text{ g}$. $\% = (0.4/0.5) \times 100 = 80\%$. (Wait, check molar masses: $K=39, KHCO_3=100$. Ratio is $62/200 = 0.31$. Yes). Correct answer: C.