$$
\begin{aligned}
& \mathrm{C}(\mathrm{~s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+400 \mathrm{~kJ} \\
& \mathrm{C}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{~g})+100 \mathrm{~kJ}
\end{aligned}
$$
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 kg of coal is burnt is
Select the correct option:
A
1600 kJ
B
3200kJ
C
4400 kJ
D
6600 kJ
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$
\mathrm{C}(\mathrm{~S})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+400 \mathrm{~kJ}
$$
1 g mole
$$
\begin{aligned}
& \mathrm{C}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{~g})+100 \mathrm{~kJ} \\
& 0.6 \times 1000=600 \mathrm{gm} \\
& 600 \times \frac{60}{100} \text { (Pure Carbon) } \\
& \quad=360 \mathrm{gm}=\frac{360}{12}=30 \text { mole (Pure Carbon) }
\end{aligned}
$$
Carbon converted into $\mathrm{CO}_2=\left(30-30 \times \frac{60}{100}\right)=12$ mole
and carbon converted in $\mathrm{CO}=30 \times \frac{60}{100}=18$ mole
Energy generated during II equation $=18 \times 100=1800 \mathrm{~kJ}$
Energy generated during $\mathrm{l}^{\text {st }}$ reaction. $=12 \times 400=4800$
Total $=1800+4800=6600 \mathrm{~kJ}$
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