2.5 gm of air dried coal sample was taken in a silica crucible, after heating it in an electric oven at $1050^{\circ} C -1100 ^{\circ}C$ for 1 hour, the residue was weighed 2.410 gm. The residue was heated in silica crucible covered with vented lid at temperature $950-9700 ^{\circ}C$ for exactly 7 minutes. After cooling the weight of residue was found to be 1.78 gm. The residue was ignited at $7500 ^{\circ}C$ to a constant weight of 0.246 gm. Calculate results of proximate analysis.

Subject:- Applied Chemistry 2.

Topic:-

Difficulty:- Medium

Solution: Weight of coal taken = 2.5 gm Mass of moisture in coal = 2.5 – 2.415 = 0.085 gm % of moisture = loss in wt. of coal/weight of coal taken x 100 = 0.085 / 2.5 x100 = 3.4 % Mass of volatile matter = 2.415 – 1.78 = 0.635gm % of volatile matter = 0.635 /2.5 x100 = 25.4% Mass of residue after ignition = 0.246 gm % of Ash = weight of residue left/weight of coal taken x100 = 0.246/2.5 x 100 = 9.8 % % of fixed Carbon = 100- (%moisture +VM+Ash) = 100-(3.4 + 25.4 + 9.8) = 100- 38.6 = 61.4% % Moisture = 3.4% % Volatile matter = 25.4% % Ash = 9.8 % % Fixed Carbon = 61.4 %