The amount of calcium oxide produced on heating 150 kg limestone (75% pure) is ______ kg. (Nearest integer)
Given : Molar mass (in g $$mol^{-1}$$) of $$Ca-40, O-16, C-12$$

The mineral limestone is almost entirely $$CaCO_3$$. When limestone is heated it undergoes thermal decomposition (calcination):

$$CaCO_3 \rightarrow CaO + CO_2$$ $$-(1)$$

Step 1 : Calculate the mass of pure $$CaCO_3$$ present in the given limestone.

Limestone given = $$150 \text{ kg}$$, but it is only $$75\%$$ pure.
Mass of pure $$CaCO_3 = 150 \times \frac{75}{100} = 112.5 \text{ kg}$$ $$-(2)$$

Step 2 : Use stoichiometry of reaction $$(1)$$.

Molar mass data:
$$CaCO_3 : 40 + 12 + 3 \times 16 = 100 \text{ g mol}^{-1}$$
$$CaO : 40 + 16 = 56 \text{ g mol}^{-1}$$

From the balanced equation $$(1)$$, $$100 \text{ g}$$ of $$CaCO_3$$ produce $$56 \text{ g}$$ of $$CaO$$.

Therefore, the mass ratio is
$$\frac{\text{mass of }CaO}{\text{mass of }CaCO_3} = \frac{56}{100} = 0.56$$ $$-(3)$$

Step 3 : Find the mass of $$CaO$$ obtained from the pure $$CaCO_3$$ in $$2$$.

Mass of $$CaO = 112.5 \times 0.56 = 63.0 \text{ kg}$$ $$-(4)$$

Hence, the amount of calcium oxide produced (nearest integer) = 63 kg.