"$$ \begin{aligned} &\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+400 \mathrm{~kJ} \\ &\mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+100 \mathrm{~kJ} \end{aligned} $$ 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 '$\mathrm{CO}_{2}$'. The heat generated when $0.6 \mathrm{~kg}$ of coal is burnt is _________."

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Accepted Answer

"Weight of coal $=0.6 \mathrm{~kg}=600 \mathrm{gm}$

$\therefore 60 \%$ of it is carbon

So weight of carbon $=600 \times \frac{60}{100}=360 \mathrm{~g}$

$\therefore$ moles of carbon $=\frac{360}{12}=30$ moles

$$ \begin{gathered} \underset{12 \text { moles }}{\mathrm{C}}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2} \\ \underset{\substack{18 \text { moles } \\ \text { (60\% of total Carbon })}}{\mathrm{C}}+\frac{1}{2} \mathrm{O}_{2} \longrightarrow \mathrm{CO} \end{gathered} $$

$\therefore$ Heat generated $=12 \times 400+18 \times 100=6600 \mathrm{~kJ}$"

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