Liquid to Gas Combustion ```Name: Romanov L. Status: student Grade: 9-12 Location: TX Country: N/A Date: May 2006 ``` Question: I have a volume/combustion question. If you have one cubic centimeter of liquid gasoline, what would be the volume of the gaseous products after a combustion? How did you formulate your answer? Replies: First a balanced equation would have to be written, which I will get by assuming fully saturated octane to represent what the major component of gasoline is, also I will assume complete combustion. C8H18 + O2 ----> CO2 + H2O Now the equation must be balanced so that the same number of C's, H's and O's are on both sides of the equation: C8H18 + 12.5 O2 ----> 9 H2O + 8 CO2 This tells us that for one molecule of gasoline, we get 9 molecules of water and 8 molecules of CO2. The volume of 1 mole of a gas at STP (standard temperature and pressure) is 22.4 L/mole. If we had 1 mole of octane, we would get 9 + 8 = 17 moles of gaseous products after complete combustion, which would be 17 mole x 22.4 L/mole = 380.8 L of gaseous products. But we have 1 mL, not 1 mole, so the next thing to do is to figure out how many moles of octane we really do have. The formula weight of octane is 114.2 and its density is 0.703 g/mL. If we have 1 mL, then we have 0.703 g, which is 0.703 mole/114.2 g/mole = 0.00616 moles. This means that 1 mL of octane = 0.00616 moles x 380.8 L/mole = 2.34 L of gaseous products would result from the combustion of octane. The correlation to gasoline is not directly applicable, however. Gasoline is a mixture of many many different products and depending on whether you get 87 octane or 93 octane, the ratio of products changes. It also matters how old the gasoline is because it has a lot of volatile components that will evaporate and change the mixture yet again. The density of gasoline, from a quick search is listed as 0.72 g/mL, but without an average molecular weight, it gives you only part of the information you are searching for. If you can find the average molecular weight of gasoline, then you can simply work through the problem above, just replace the FW and density figures. But again, you are also assuming that the balanced equation is applicable as well. I think that to get a "real" answer one would need to actually have a gasoline sample and quantitate the component ratios, then write a balanced equation for each and weight the volume of each component by the component ratio. Everything else will just be a close estimate. Matt Voss This question can have several answers depending upon how you formulate the question, but the principles involved are these: Assume gasoline is octane (C8H18). This is an oversimplification because gasoline is a mixture of various hydrocarbons. The general approach is to make the calculation based on 1 mol of octane, then multiply the result by the number of mols of octane. Save this for last. A good approximation is that the volume of any liquid reactant or product is essentially zero compared to the volume of gas of the same reactant or product. It is also a good approximation to assume the ideal gas law, that is: P*V = N*R*T, where P is the pressure in atmospheres, V is the volume in liters, N is the number of moles of the particular GAS, T is in kelvins, and R = 0.08205 liter atm / mol kelvin. The balanced chemical reaction of octane with oxygen is: C8H18 + (25/2) O2 ---> 8 CO2 + 9 H2O Other parameters need to be specified: Do you want to consider H2O to be a gas or a liquid? It is evident that how you specify that physical state greatly affects the change in the number of mols of gaseous products and reactants. What temperature(s) do you want to set for the reactant O2 and the products CO2 and H2O. If all are at 298 kelvins the change in volume will be very different from the value you get if you consider O2 to be at 298 kelvins and the products to be at 1000 kelvins. The customary way is to assume the temperature to be a constant 298 kelvins and the water to be a liquid. Then, on a per mole C8H18 basis the change in the number of moles of gases is: (8 - 25/2) = - 4.5. In practice, of course, that is not how an internal combustion engine works. You are free to do the calculation considering the product H2O to be a gas, as well as the CO2, and you are free to choose the temperature to be whatever you desire. You could also choose the volume of O2 to be 298 kelvins and the products to be at some high temperature, say 1000 kelvins. Various selections cause major changes in the change in volume. The key principles are to remember that it is the change in the number of MOLS of GAS (under whatever conditions you choose for each) that determines the change in volume. The volume of liquids are negligible by comparison in all cases. Finally, octane has a molecular weight of 114.2 gm/mol and a liquid density at 25 C = 298 kelvins of about 0.7 gm / cm^3. So the 1 cm^3 of octane is about 0.006 mols of octane in 1 cm^3. Save this "scale" factor until the end, so you do not have to keep converting values to that amount of octane. You only need to do it once. Vince Calder Romanov, You would first apply some stoichiometry and then gas laws. Stoichiometry: 1) write the balanced equation (this will be a problem since gasoline is a complex combination of compounds - you may have to choose the most abundant liquid in gasoline) you will have to assume perfect combustion (to make it easier) 2) convert the volume of gasoline to mass 3) calculate the mass of gaseous products (be sure you know what temperature you are working in so that you can decide if the water produce is gaseous or not). Ideal Gas: 1) Assume that all the gases in the system is an ideal gas 2) Calculate the volume of gas produced given a certain pressure and temperature (use V=nRT/P) Greg (Roberto Gregorius) Click here to return to the Chemistry Archives

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