Write the decomposition reaction

Limestone is mainly calcium carbonate, $CaCO_3$. On heating: $CaCO_3 \rightarrow CaO + CO_2$

Find the mass of pure limestone

Given limestone is $75\%$ pure.

Total mass of sample $=150\,\text{kg}$

So, mass of pure $CaCO_3$ is: $150 \times \frac{75}{100}=112.5\,\text{kg}$

Calculate molar masses

$M(CaCO_3)=40+12+3(16)=100\,\text{g mol}^{-1}$ $M(CaO)=40+16=56\,\text{g mol}^{-1}$

Use stoichiometry

From the reaction: $100\,\text{g of } CaCO_3 \text{ gives } 56\,\text{g of } CaO$

Therefore, $112.5\,\text{kg of } CaCO_3 \text{ gives } 112.5\times \frac{56}{100}$ $=63\,\text{kg}$

Final answer

The amount of calcium oxide produced is: $\boxed{63\,\text{kg}}$

Comparison with stored correct answer

Stored correct answer = $63$

My derived answer also is $63$, so they agree.