50 grams of lime stone on calcination liberated 16.5 grams of carbon dioxide. Percentage yield of carbon dioxide is equal to

$$\begin{gathered} {\mathrm{CaCO}}_3\left(\mathrm{s}\right) \stackrel{∆}{\to } \mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right); \\ 100 \mathrm{g} 44 \mathrm{g} \\ 50 \mathrm{g} 22 \mathrm{g} ; \\ \mathrm{If} \mathrm{the} \mathrm{yield} \mathrm{is} 100\% \mathrm{then} \mathrm{the} \mathrm{weight} \mathrm{of} {\mathrm{CO}}_2 \mathrm{liberated} =22 \mathrm{g} ; \\ \mathrm{but} \mathrm{the} \mathrm{weight} \mathrm{of} {\mathrm{CO}}_2 \mathrm{obtained} \mathrm{is} 16.5 \mathrm{g}; \\ \mathrm{Thus} \mathrm{percentage} \mathrm{yield} \mathrm{of} {\mathrm{CO}}_2=\frac{16.5}{22}\times 100=75\% ; \end{gathered}$$