(i) weight of pure coal = \(\frac{80}{100}\)x 10 = 8.0 kg
Weight of coal converted into CO2 = 8 x \(\frac{60}{100}\)
= 4.8 kg
Weight of coal converted in CO = 8 − 4.8
= 3.2 kg
Moles of C in 4.8 kg coal
= \(\frac{4800}{12}\) = 400 moles
Moles of C in 3.2 kg coal
= \(\frac{3200}{12}\) = 266.67 moles
Now C + O2 → CO2 (g) + 394 kJ
1 mole of C gives heat = 394 kJ
\(\therefore\) 400 moles give heat = 400 × 394
= 157600 kJ
C + \(\frac{1}{2}\)O2 → CO + 111 kJ
1 mole of C gives heat = 111 kJ
\(\therefore\) \(\frac{3200}{12}\) mole of C gives heat
= 111 × \(\frac{3200}{12}\) = 29600 kJ
Total heat generated = 157600+29600
= 187200 kJ
(ii) C + O2 → CO2 + 394
1 mole of C gives heat = 394 kJ
= \(\frac{8000}{12}\) mole of C gives heat
= 394 x \(\frac{8000}{12}\) = 262667 kJ
(iii) Heat lost due to inefficient oven
= 262667 − 187200 = 75467 kJ
\(\therefore\) % loss = \(\frac{75467}{262667}\) x 100 = 28.8%