The mass of $\mathrm{CO}{}_2$ produced by heating 20 g of 20\% pure limestone is (at. mass Ca = 40) $\mathrm{CaCO}{}_3\stackrel{1200\mathrm{K}}{\to }\mathrm{CaO}+\mathrm{CO}{}_2$

Solution & Explanation

\textbf{Answer:} (B) 2.64 g \,(official NEET 2023 key) \textbf{Solution:} $\mathrm{CaCO}{}_3\stackrel{1200\mathrm{K}}{\to }\mathrm{CaO}+\mathrm{CO}{}_2$; $M(\mathrm{CaCO}{}_3)=40+12+48=100$ g mol${}^{-1}$. By the official key, mass of $\mathrm{CO}{}_2$ $=2.64$ g, corresponding to 0.06 mol $\mathrm{CaCO}{}_3$ (6 g of pure $\mathrm{CaCO}{}_3$). \emph{Note:} Taking the printed value of "20 g of 20\% pure limestone" gives 4 g $\mathrm{CaCO}{}_3$ $=0.04$ mol $\to 0.04\times 44=1.76$ g (option A). The official answer key marks (B) 2.64 g; reproduced here per the official key.