The mass of $80\%$ pure calcium carbonate required to prepare $11.2 \mathrm{L}$ of ${\mathrm{CO}}_2$ at STP is……g

1 mole of calcium carbonate degrades to form 1 mole of calcium oxide along with carbon dioxide gas.

${\mathrm{CaCO}}_3\stackrel{∆}{\to }\mathrm{CaO}+{\mathrm{CO}}_2$

$100\mathrm{g} \mathrm{of} {\mathrm{CaCO}}_3=22.4\mathrm{L} \mathrm{of} {\mathrm{CO}}_2$

$\mathrm{xg} \mathrm{of} {\mathrm{CaCO}}_3=11.2\mathrm{L} \mathrm{of} {\mathrm{CO}}_2$

And the number of moles of ${\mathrm{CaCO}}_3=80/100$

$$\begin{gathered} \mathrm{So}, \mathrm{x}\times (80/100)\times 22.4=11.2\times 100 \\ \mathrm{x}=62.5\mathrm{g} \end{gathered}$$

Hence, $62.5 \mathrm{g}$ of calcium carbonate is required here.