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