C(s)+O₂( g) → CO₂( g)+400 kJ ; C(s)+1/2 O₂( 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 _________.

Solution

Chemistry ma’am · 3 steps

Find the moles of actual carbon first, then split them between CO and CO2 and add the heats.

1. Pure carbon only. Only 60% of the coal is carbon, so 0.6 kg of coal holds 0.6 × 600 g = 360 g of carbon. That is 360/12 = 30 mol of C.
2. Split the carbon. 60% of the carbon gives CO: 0.6 × 30 = 18 mol. The remaining 12 mol gives CO₂. Each equation gives the heat released per mole of carbon burnt.

   Product	Moles of C	Heat per mole	Heat released
   CO	18	100 kJ	1800 kJ
   CO₂	12	400 kJ	4800 kJ

3. So. Total heat = 1800 + 4800 = 6600 kJ, which is option (d).

The idea to keep. In mixed-combustion problems, convert impure mass to moles of pure carbon first. Then give each fraction its own heat per mole and add them up.

Answer from the NTA final answer key. Solution by PrepWiser.