A gas produced as a by-product from the carbonization of coal has the following composition, mole %: carbon dioxide 4, carbon monoxide 15, hydrogen 50, methane 12, ethane 2, ethylene 4, benzene 2, balance nitrogen. Using the data given in Appendix C (available online at booksite .Elsevier/Towler), calculate the gross and net calorific values of the gas. Give your answer in MJ/m3, at standard temperature and pressure.

6059.63 kcal/kg.

Explanation:

CH4 consist of Hydrogen and carbon. Therefore, Hydrogen in CH4 = 12 × 4 = 48 kg,  carbon in CH4 = 12 × 12 = 144kg.

For ethane, the amount of hydrogen present = 2 × 6 = 12kg and that of carbon in ethane = 2 × 24 = 48kg.

The weight of carbon in CO = 15 × 12 = 18kg and the weight of Hydrogen in CO = 15 × 16 = 240kg.

For hydrogen, its weight in H2 = 50 × 2 = 100kg.

For CO2, carbon has = 4 × 12 = 48 kg and oxygen has = 4 × 32 = 128kg.

For C6H6, carbon has 2 × 72 = 144kg and hydrogen has 2 × 6 = 12kg.

For N2, the amount of nitrogen= 11 × 28 = 308 kg.

For CH2= CH2, carbon has 4 × 24 = 96kg and hydrogen = 4 × 4 = 16kg.

The gross calofiric value = 1/100 [ 8080 C + 34500 + ( H - O/8) + 22405].

Where the total weight = 128 + 180+ 48 + 240 + 100 + 144 + 48 + 12 + 48 + 16 + 96 + 308 + 12 + 144 = 1524 kg.

The percentage by weight of carbon = total weight of carbon/total weight × 100.

The total Weight of carbon= 48 + 180 + 144 + 48 +144 + 96 = 660kg.

The percentage weight of carbon = 660/1524 × 100 = 43.3 %.

The percentage weight of hydrogen = total weight of hydrogen/total weight × 100.

The total weight of hydrogen = 100 + 12 + 48 + 16 +12 = 188.

The percentage weight of Hydrogen = 188/ 1524 × 100 = 12.33%.

Percentage weight of oxygen = total weight of oxygen/total weight × 100.

Percentage weight of oxygen = ( 128 + 240) / 1524 × 100 = 24.15%.

The gross calorific value = 1/100 [ 8080 × 43.3 ,+ 34500 ( 13.33 - 24.15/8) ] = 6711.02 kcal/kg.

Net calorific value = 6711.02 - 0.09 × 12.3 × 587 = 6059.63kcal/kg.

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.

they both have a y in them