The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane (C8H18) and calculate the mass (in kg) of carbon dioxide that is added to the atmosphere per 10.0 kg of octane burned. (Hint: Begin by writing a balanced equation for the combustion reaction.)

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All right. Hi, everyone. So this question says that the combustion of gasoline produces carbon dioxide and water assume gasoline to be pure pentane C five H 12 and calculate the mass in kilograms of carbon dioxide that is added to the atmosphere per 1.0 kg of pentane burnt. Hint begin by writing a balanced equation for the combustion reaction. And here we have four different answer choices labeled A through D. So first let's go ahead and get started by writing out the balanced equation or the combustion reaction described recalled combustion is the reaction of a hydrocarbon. In this case, pentane with oxygen that produces water vapor and carbon dioxide. So first, we have pentane thats C five H 12, which is a guess reacting with oxygen gas to produce carbon dioxide and water vapor. All right. So in the case of a combustion reaction, it's always best to start out by balancing carbon, followed by hydrogen and then oxygen. The reason for this is because oxygen is present in both of our products, which is going to make it more difficult to balance in an earlier step. Right. So first starting off with carbon there are five carbons on the reactant side and only one carbon from carbon dioxide on the product side to balance this, I'm going to add a coefficient of five in front of CO two. And this balances out carbon, I have five on each side of the reaction. So up next, we have hydrogen on the react on the reactant side, we have 12 hydrogen atoms from pentane and two from water vapor. So because one molecule of water vapor contains two hydrogen atoms, I'm going to balance out hydrogens on either side by adding a coefficient of six in front of H2O. This brings up the number of hydrogen. Excuse me, the number of hydrogens on the right side, the 12 meaning that hydrogen is now balanced. On the right side, I wrote water not hydrogen, but our last step is to balance out oxygen. Now, on the reactant side, there are only two oxygen atoms. Whereas on the right side, there are 10 oxygen atoms in five molecules of carbon dioxide and an additional six from six molecules of water vapor. So this means that there are a total of 16 oxygen atoms on the right side of the reaction. So to balance out the number of oxygens, I'm going to add a coefficient of eight in front of 02 to produce 16 oxygen atoms on either side of the reaction. And so now we have a fully balanced reaction equation for the combustion process. So using the ratios between reactants and products. In this combustion process, we can convert the mass of pine. In this case, 1 kg into the mass of carbon dioxide produced. But recall that these coefficients in the balanced chemical equation describe the ratio or the ratios of moles, right. So to convert mass into moles, we're going to have to use the molar masses of both pentane and carbon dioxide. Now, the molar mass of pentane is equal to 72.15 g per mole. Whereas the molar mass of carbon dioxide is equal to 44.01 g per mole. So starting off with our given, we're going to burn 1 kg or 1.0 kg of PTE. Now, our first step would normally be to convert our mass into moles using the molar mass, right? But because the mass or the initial mass is provided in units of kilograms, our first step is actually to convert kilograms into grams before using the molar mass. So here recall that 1 kg is equal to 1000 g. And so this cancels out kilograms and gives us a value in the units of grams of Pake, which we can then use to find the number of moles of pentane in 1.0 kg. Now recall that the idea is always to cancel the starting units. So in applying the molar mass of pentane as a conversion factor, 72.15 g should go in the denominator. Whereas one mole of panting shed go on the numerator. This ensures that grams of pentane cancel out, leaving me with moles of plantain in my answer at this point. So now that we have that we can use the coefficients of the balanced chemical reaction to relate the moles of pentane to the moles of carbon dioxide. So once again, moles of pentane should go on the denominator to make sure that cancels out. So according to the balance reaction, there are five moles of carbon dioxide and one mole of pine and that cancels out malls of painting. And so at this point, our answer is in units of moles of carbon dioxide, meaning that we're going to apply the molar mass of carbon dioxide to convert moles into grams. So that's 44.01 g of carbon dioxide in the numerator over one mole of carbon dioxide in the denominator, canceling out moles of carbon dioxide. So after evaluating this entire expression, the mass of carbon dioxide produced in grams is 3049 0.9 grams of carbon dioxide. But because our answers are expressed in units of kilograms, there is one more conversion we have to make. So once again, recall that 1 kg is equivalent to 1000 g, meaning that were going to take our value in grams of carbon dioxide and divide by 1000. This yields 3.0499 kg of co two, which we can then express as 3.0 kilograms of co two after rounding to two significant figures and there you have it. So our answer is 3.0 kg of carbon dioxide which corresponds to option C in the multiple choice. So with that being said, thank you so very much for watching and I hope you found this helpful.

The combustion of gasoline produces carbon dioxide and water. Assume gasoline to be pure octane (C8H18) and calculate the mass (in kg) of carbon dioxide that is added to the atmosphere per 10.0 kg of octane burned. (Hint: Begin by writing a balanced equation for the combustion reaction.)

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Key Concepts

Combustion Reaction

Stoichiometry

Molar Mass