$$C(s) + O_2(g) \to CO_2(g) + 400$$ kJ
$$C(s) + \frac{1}{2}O_2(g) \to CO(g) + 100$$ kJ
When coal of purity 60% is allowed to burn in presence of insufficient oxygen, 60% of carbon is converted into 'CO' and the remaining is converted into '$$CO_2$$'. The heat generated when 0.6 kg of coal is burnt is

We are given: $$C(s) + O_2(g) \to CO_2(g) + 400 \text{ kJ}$$ $$C(s) + \frac{1}{2}O_2(g) \to CO(g) + 100 \text{ kJ}$$

The coal has a purity of 60%, meaning 60% of the coal is carbon. The mass of coal is 0.6 kg = 600 g. So the mass of carbon is $$0.60 \times 600 = 360$$ g. The number of moles of carbon is $$\frac{360}{12} = 30$$ mol.

Now, 60% of the carbon is converted to CO and the remaining 40% is converted to $$CO_2$$.

Moles of C converted to CO: $$0.60 \times 30 = 18$$ mol. Each mole releases 100 kJ, so heat from CO formation = $$18 \times 100 = 1800$$ kJ.

Moles of C converted to $$CO_2$$: $$0.40 \times 30 = 12$$ mol. Each mole releases 400 kJ, so heat from $$CO_2$$ formation = $$12 \times 400 = 4800$$ kJ.

Total heat generated = $$1800 + 4800 = 6600$$ kJ.

Hence, the correct answer is Option D.