Answer is: 63
\(\underset{150\ kg}{CaCO_3} \rightarrow CaO + CO_2\)
\(\text{Pure} \ CaCO_3 \text{ is 150 kg} \ \times \ \frac{75}{100}\)
\(= 150 \times 10^3 \times \frac{75}{100} = 1125 \times 10^2 gm\)
\(\text{Mole of } CaCO_3 = \frac{1125 \times 10^2}{100} = 1125 \text{mole}\)
\(\text{Mole of CaO} = 1125 \ mole\)
\(\text{Mass of CaO}= 1125 \times 6\)
\(=63000 \ gm\)
\(= 63 \ kg\)