Chemistry - Mole Concept Question with Solution | TestHub

$\mathrm{C}\left(\mathrm{S}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right)+400 \mathrm{kJ}$

$1$ mole

$\mathrm{C}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to \mathrm{CO}\left(\mathrm{g}\right)+100 \mathrm{kJ} ...\left(2\right)$

$0.6\mathrm{Kg}\times 1000$

$=600\mathrm{gm}$

Mass of $60\%$ pure coal

=$600\times \frac{60}{100}$ (Pure Carbon)

$$\begin{gathered} =360\mathrm{gm} \\ \mathrm{Number} \mathrm{of} \mathrm{moles}=\frac{360}{12}=30 \mathrm{moles} \mathrm{of} \mathrm{pure} \mathrm{carbon} \end{gathered}$$

Moles of 60% pure Carbon converted into ${\mathrm{CO}}_2=\left(30-30\times \frac{60}{100}\right)$

$=12$ mole and

Moles of 60% pure carbon converted in to $\mathrm{CO}=30\times \frac{60}{100}=18$ mole

Energy generated during $2$ equation $=18\times 100$

$=1800 \mathrm{kJ}$

Energy generated during $1^{\mathrm{st}}$ reaction.

$=12\times 400$

$=4800$

Total $=1800+4800=6600 \mathrm{kJ}$