Saved Bookmarks
| 1. |
C(s) + O2(g) → CO2(g) + 400 kJC(s) + 1/2 O2(g) → CO(g) + 100 kJWhen 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 ______. (A) 1600 kJ (B) 3200 kJ (C) 4400 kJ (D) 6600 kJ |
|
Answer» Correct option is (D) 6600 kJ C(S) + O2(g) → CO2(g) + 400 kJ 1 g mole C(s) + 1/2 O2 (g) → CO (g) + 100kJ ..... 2 (II) 0.6 x 1000 = 600 gm 600 x 60/100 (Pure Carbon) = 360 gm = 360/12 = 30 mole (Pure Carbon) Carbon converted into CO2 = (30 - 30 x 60/100) = 12 mole and carbon converted in CO = 30 x 60/100 = 18 mole Energy generated during II equation = 18 x 100 = 1800 kJ Energy generated during 1st reaction. = 12 x 400 = 4800 Total = 1800 + 4800 = 6600 kJ |
|