Page:International Library of Technology, Volume 93.djvu/83

 23 percent., by weight, of air is O. The combustion of 1 pound of C may be represented as follows:

Mixture, Elements, Products, in Pounds in Pounds in Pounds Carbon, 1 Carbon, 11 = Carbon dioxide, 3.67 Air 11.6 = Oxygen, 2.67 Nitrogen, 8.93 Nitrogen, 8.93 Total, 12.6 Total, 12.6 Total, 12.6

That is, 1 pound of C requires for its complete combustion 11.6 pounds of air. Of this air, 2.67 pounds is O, which combines with the pound of C, forming 3.67 pounds of CO2. The 8.93 pounds of nitrogen contained in the air passes off with the CO, and takes no part in the combustion.

Take, next, the complete combustion of 1 pound of hydrogen. The product of the combustion is water, H2O. It has been shown that H20 is composed, by weight, of 2 parts of H to 16 parts of O. Hence, 1 pound of H requires 16 ÷ 2 = 8 pounds of O to unite with it. The air required to furnish 8 pounds of O is 8 ÷ .23 = 34.8 pounds. The process of combustion is, therefore, as follows:

Mixture, Elements, Products, in Pounds in Pounds in Pounds Hydrogen, 1 Hydrogen, 1 = Water 9 Air 34.8 = Oxygen, 8 Nitrogen, 26.8 Nitrogen, 26.8 Total, 35.8 Total, 35.8 Total, 35.8

16. There is one other case that may occur; the combustion of C may not be complete. If insufficient air or is supplied to the burning C, it is possible for the C and to form another gas, carbon monoxide, CO, instead of carbon dioxide, CO2. The combustion of 1 pound of C to form CO, of course, requires only one-half of the O that would be necessary to form CO a, because, in forming CO, one atom of C unites with one atom of O instead of two. To burn 1 pound of C to CO2 requires 11.6 pounds of air. To burn it to CO will,