The correct value for the mass of $\mathrm{CO}_2$ gas produced by heating $50 \mathrm{~g}$ of a lime

Question

The correct value for the mass of $\mathrm{CO}_2$ gas produced by heating $50 \mathrm{~g}$ of a limestone sample that is $80\%$ pure is (Atomic mass of $\mathrm{Ca}=40, \mathrm{C}=12, \mathrm{O}=16$)

Professor PC Thomas Classes · Answer

(a) : The chemical equation for the thermal decomposition is $\mathrm{CaCO}_3 \xrightarrow{\Delta} \mathrm{CaO} + \mathrm{CO}_2 \uparrow$. The molar mass of $\mathrm{CaCO}_3$ is $100 \mathrm{~g/mol}$. The mass of pure $\mathrm{CaCO}_3$ in the $50 \mathrm{~g}$ sample is $80\%$ of $50 \mathrm{~g} = \frac{80}{100} 	imes 50 = 40 \mathrm{~g}$. According to the stoichiometry, $100 \mathrm{~g}$ of pure $\mathrm{CaCO}_3$ yields $44 \mathrm{~g}$ of $\mathrm{CO}_2$. Therefore, $40 \mathrm{~g}$ of pure $\mathrm{CaCO}_3$ will yield $\frac{44}{100} 	imes 40 = 17.6 \mathrm{~g}$ of $\mathrm{CO}_2$.