In an oven, due to insufficient supply of oxygen,60% of the carbon is converted to carbon dioxide
whereas the remaining 40% is converted into carbon monoxide. If the heat of combustion of carbon to CO2 is 394 kJ mol-1 while that of its
oxidation to CO is 111 kJ mol-1,
calculate the total heat produced in the oven by burning 10 kg
of coal containing 80% carbon by weight.
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Answer:
(i) weight of pure coal = 8010080100x 10 = 8.0 kg Weight of coal converted into CO2 = 8 x 6010060100 = 4.8 kg Weight of coal converted in CO = 8 − 4.8 = 3.2 kg Moles of C in 4.8 kg coal = 480012480012 = 400 moles Moles of C in 3.2 kg coal = 320012320012 = 266.67 moles Now C + O2 → CO2 (g) + 394 kJ 1 mole of C gives heat = 394 kJ ∴∴ 400 moles give heat = 400 × 394 = 157600 kJ C + 1212O2 → CO + 111 kJ 1 mole of C gives heat = 111 kJ ∴∴ 320012320012 mole of C gives heat = 111 × 320012320012 = 29600 kJ Total heat generated = 157600+29600 = 187200 kJ (ii) C + O2 → CO2 + 394 1 mole of C gives heat = 394 kJ = 800012800012 mole of C gives heat = 394 x 800012800012 = 262667 kJ (iii) Heat lost due to inefficient oven = 262667 − 187200 = 75467 kJ ∴∴ % loss = 7546726266775467262667 x 100 = 28.8%
Answer:
(i) weight of pure coal = 8010080100x 10 = 8.0 kg Weight of coal converted into CO2 = 8 x 6010060100 = 4.8 kg Weight of coal converted in CO = 8 − 4.8 = 3.2 kg Moles of C in 4.8 kg coal = 480012480012 = 400 moles Moles of C in 3.2 kg coal = 320012320012 = 266.67 moles Now C + O2 → CO2 (g) + 394 kJ 1 mole of C gives heat = 394 kJ ∴∴ 400 moles give heat = 400 × 394 = 157600 kJ C + 1212O2 → CO + 111 kJ 1 mole of C gives heat = 111 kJ ∴∴ 320012320012 mole of C gives heat = 111 × 320012320012 = 29600 kJ Total heat generated = 157600+29600 = 187200 kJ (ii) C + O2 → CO2 + 394 1 mole of C gives heat = 394 kJ = 800012800012 mole of C gives heat = 394 x 800012800012 = 262667 kJ (iii) Heat lost due to inefficient oven = 262667 − 187200 = 75467 kJ ∴∴ % loss = 7546726266775467262667 x 100 = 28.8%