How much amount of ${\mathrm{CaCO}}_3$ in gram (having 50 percent purity) produce 0.56 liter of ${\mathrm{CO}}_2$ at STP on heating?

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

To produce 0.56 liter or $\frac{0.56}{22.4\times 10}=\frac{1}{40}\text{ mole }$

$$\begin{gathered} \text{ Amount of }{\mathrm{CaCO}}_3\text{ required to produce }\frac{1}{40}\text{ mole of C}{O}_2 \\ =\frac{1}{40}\times 100=\frac{5}{2}\text{ gram } \end{gathered}$$

Percentage purity is = 50%

$\text{ So, amount required }=2\times \frac{5}{2}=5\text{ gram }$