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
Question
A $0.5 \text{ g}$ sample of $KHCO_3$ and $K_2CO_3$ was heated. The loss in weight was $0.031 \text{ g}$. The percentage of $K_2CO_3$ in the mixture is:
Answer: A
💡 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.
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