written 6.3 years ago by
vibhavarikulkarni
• 480
|
modified 2.2 years ago
by
pedsangini276
• 4.7k
|
Subject:- Applied Chemistry 2.
Topic:- Fuels.
Difficulty:- Medium
|
written 6.2 years ago by
vibhavarikulkarni
• 480
|
modified 2.2 years ago
by
yashbeer
★ 11k
|
Solution :
C + O2 CO2 C = 85% = 0.85 kg
12 + 32 44 H = 10% = 0.10 kg
H2 +1/2O2 H2O 5% = 0.05 kg 5% = 0.05 kg
2 + 16 18
| Element = weight in kg | Weight of O2 required for complete combustion in kg |
|---|---|
| C = 0.85 | 0.85 x 32/12 = 2.26 |
| H =0.1 | 0.1 x 8 =0.8 kg |
| O = 0.05 | - |
| Total oxygen = 3.06 kg |
Wt. of oxygen = wt. of oxygen needed – wt. of oxygen present = 3.06 – 0.05 = 3.01 kg
Wt. of air required for complete combustion = [Air contains 23%O2 by weight] 3.01 x 100/23 = 13.08 kg per 1 kg of coal Air required for 5 kg of coal =13.08 x 5 = 65.40 kg Volume of air 28.94 kg of air = 22,400 ml volume at NTP 65.4kg of air = 22400 x 65.4/28.94 = 50620.6 ml of air Answer: Weight of air required= 65.40 kg Volume of air= 50620.6 ml of air
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