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C(s) + O2 (g) → CO2 (g) + 394 kJ

C(s) + \(\frac{1}{2}\)O2 (g) → CO (g) + 111 kJ

(i) In an oven using coal (assume the coal is 80% carbon in weight), insufficient oxygen is supplied such that 60% of carbon is converted to CO2 and 40% carbon is converted to CO. Find out the heat generated when 10 kg of coal is burnt in this fashion.

(ii) Calculate the heat generated if a more efficient oven is used so that only CO2 is formed.

(iii) Calculate the percentage loss in heating value for the inefficient oven.

1 Answer

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(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%

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