Verified Solution • 2025–26 Session moderate

QJEEM22Ch7dee: JEE Main 2022 Chemistry - Class 11 JEEM22Ch7dee Solution: C ( s ) + O 2 (   g ) → CO 2 (   g ) + 400   kJ…

C(s)+O2( g)→CO2( g)+400 kJC(s)+12O2( g)→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 'CO2'. The heat generated when 0.6 kg of coal is burnt is _________.

Multiple Choice Options

  • A 1600 kJ
  • B 3200 kJ
  • C 4400 kJ
  • D 6600 kJ

Step-by-Step Verified Solution

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

∴60%\therefore 60 \% of it is carbon

So weight of carbon =600×60100=360 g=600 \times \frac{60}{100}=360 \mathrm{~g}

∴\therefore moles of carbon =36012=30=\frac{360}{12}=30 moles

C12 moles +O2⟶CO2C18 moles  (60\% of total Carbon )+12O2⟶CO

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

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