Calculate the weight and volume of air required for complete combustion of 5kg coal with following composition, C = 85%, H = 10 %, O= 5%.

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written 6.9 years ago by

vibhavarikulkarni • 510

modified 2.8 years ago by

pedsangini276 • 4.8k

Subject:- Applied Chemistry 2.

Topic:- Fuels.

Difficulty:- Medium

advanced engineering chemistry

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written 6.8 years ago by

vibhavarikulkarni • 510

modified 2.8 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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