Every combustion reaction involves the reaction of molecular oxygen with some organic fuel. Whether the fuel is methane, propane, diesel or something else, the products of the reaction will be carbon dioxide and water. This means that we can write the equation for any combustion reaction, and by balancing the elements on each side, determine the amount of oxygen required to completely burn the fuel.
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Things you need
- Periodic Table
Write the equation for the combustion reaction. The equation should place one molecule of fuel and one molecule of oxygen on the left-hand side, since they are reactants. The products, carbon dioxide and water, should go on the right-hand side. For example, your equation might show C3H8 + O2 --> CO2 + H20.
Balance the equation by writing the necessary coefficients in front of each molecule. Because elements are neither created nor destroyed in a reaction, the quantity of each element on the left must equal the quantity on the right. Our original equation, C3H8 + O2 --> CO2 + H20, would become C3H8 + 5O2 --> 3CO2 + 4H20.
Calculate the molecular weights of the reactants. The balanced equation shows the smallest-scale representation of the combustion process. Therefore, if you calculate the weight of fuel and oxygen shown in the equation, that gives you the ratio of weights required in the actual reaction. In the equation C3H8 + 5O2 --> 3O2 + 4H20, there is one molecule of the fuel and five molecules of oxygen. The fuel weighs three times the atomic weight of carbon plus eight times the atomic weight of hydrogen, or approximately 3 * 12 + 8 * 1 = 44 atomic mass units. The five units of oxygen collectively weigh 5 * (16 * 2) = 160 atomic mass units.
Calculate the air-to-fuel ratio. Note that air is only 23.2 per cent oxygen by mass, so the air-to-fuel ratio will be much larger than the oxygen-to-fuel ratio. To get the required weight of air, divide the amount of oxygen by 0.232. A reaction that uses 160 atomic mass units of oxygen will require 160 / 0.232 = 689.66 units of air. The air-to-fuel ratio would then be 689.66 / 44 = 15.7. This is the ratio of masses, not volumes.
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