Isooctane, C8H18, is the component of gasoline from which
the te...

Question

Isooctane, C8H18, is the component of gasoline from which 
the term octane rating derives. 
(b) Assuming that gasoline is 100% isooctane, that isooctane 
burns to produce only CO2 and H2O, and that 
the density of isooctane is 0.792 g/mL, what mass of 
CO2 in kilograms is produced each year by the annual 
U.S. gasoline consumption of 4.6 * 1010 L?

Accepted Answer

Welcome back everyone, we're told that decade is a component of diesel, assuming diesel is 100% decade. The complete combustion of decade produces carbon dioxide and water and the density of decade is 1000.73 g per mil leader, we need to calculate the mass of carbon dioxide and kilograms produced if the yearly diesel consumption in the U. S. Is 1.78 times 10 to the 11th power Leaders. Our first step is to write out a chemical equation. So according to our prompt, we have our deck gain C 10 and let's write this neatly. So C 10 H 22 And it's reacting with oxygen in a combustion reaction. So we have 02 and we're going to be producing our products because this is a combustion. We know that our product should be carbon dioxide and water. And so now we need to make sure that this is balanced. So writing out our atoms, we have carbon, sorry carbon, hydrogen and oxygen On the product side, we have carbon hydrogen and oxygen counting our atoms on the reactant side, we have 10 atoms of carbon On the react inside we have 22 atoms of hydrogen. And on the reactive side we have two atoms of sorry, 22 atoms of hydrogen and two atoms of oxygen. So now on our product side counting our atoms, we have one atom of carbon, we have a total of three atoms of oxygen and for our hydrogen we can count a total of two hydrogen. So to mend our imbalance on both sides of the equation, we're going to need to add coefficients and so and sorry, this should be 22 here. So to bounce things out, we're gonna go ahead and place a coefficient of two in front of our decade, Which is going to change our carbon to now 20 atoms on the reactant side. And our hydrogen on the reactant side to 44 Where to mend this on the product side, we're going to place a coefficient of 20 On the product side, in front of carbon dioxide, which is going to give us now 22 atoms of carbon on the product side, a total of 41 atoms of oxygen on the product side. And now we still need to bounce out our hydrogen. So to fix hydrogen on the product side, we're going to place a coefficient of 22 in front of our water, which is now going to give us a total of also Atoms of hydrogen on the product side. So now our hydrogen is balanced but our oxygen will now change to now 62 atoms of oxygen where on our reactant sides to bounce out oxygen, we're just going to place a coefficient of 31 in front of our oxygen. And so this is now going to give us a total of 62 oxygen's on the reactant side. And so just to go over our carbon again on the product side, adding that coefficient of 20 on the product side should have actually been account of just 20 carbon atoms. And so now we can see that based on our added coefficients, we now have a balanced equation. Now we want to take note of our molar masses of our reactant specifically are re agents that are discussed in the prompt, which is our deck Ain. And then our mass of carbon dioxide, which we need to find in kilograms for our final answer. So beginning with our molar mass of decade From our periodic table, we will see that it corresponds to a mass of 142.29 g per mole. And now for our molar mass of carbon dioxide, we will see on our product table that it corresponds to a molar mass of 44.01 g per mole. Now that we have these molar masses noted down, We need to get to our final answer by using stock geometry. So finding our mass of C. 02 produced, We're going to begin our calculation with the mass of our decade. We're going to need to refer to the info in our prompt, giving us our assumption that our diesel is 100% deck ain and we're given our density of decade as well as the volume of diesel produced, which is ultimately our volume of decade. So we're going to recall our density equation where density is equal to mass over volume. And so to isolate for mass of our decking. C- 10 age 22. We would calculate it by taking our density of decking or diesel times the volume. And so plugging in what we know from the prompt, we're going to get that density of 0.73 g per mil leader. Now we want our final unit to just be grams. So we're going to multiply by the conversion factor to go from milliliters in the numerator to leaders in the denominator where then we will plug in our volume and let's scoot this over. So we have enough room. We're going to plug in our volume which is given in the prompt for diesel, which we are assuming is 100% decade as 1.78 times to the 11th Power Leaders. And so this allows us to not only cancel out leaders but Mil leaders. When we recall that we have an equivalent for one millimeter of 10 to the negative third power of our base unit leader. So plugging everything carefully into our calculators, we're going to get a mass equal to 1. Times 10 to the negative. And rather it should be positive 14. So to the 10th of the positive 14 power and we have units of grams of our decking here. So now we're going to begin our study geometry with this calculation here for our massive decade. So beginning with 1.299 times 10 to the 14th grams of C 10 H 22 we're going to cancel out grams of decade by going from grams of decades in the denominator. So sorry, C 10 H 22 2 moles of decade in the numerator where we would recognize from our molar mass, we have an equivalent of 142.29 g of decade for one mole of decade. Now we can get rid of grams of decade and focus on going from moles of decade to are moles of carbon dioxide by utilizing our molar ratio from our balanced equation as our next conversion factor. Where according to our balanced equation, we have a ratio between moles of decade. So moles of decade in the denominator, Two moles of our carbon dioxide in the numerator. According to our balanced equation, we have two moles of decades for 20 moles of carbon dioxide. So we're using that as a conversion factor. So in our numerator we have 20 moles of carbon dioxide for two moles of decades. We can get rid of moles of decade now and now we want to end up with grams of carbon dioxide as our final unit where actually according to the prompt, the final unit should be in kilograms. So we're gonna have to do another conversion. But first making room for our next conversion which will continue here below. We're multiplying to go from moles of carbon dioxide in the denominator two g of carbon dioxide by plugging in that molar mass of carbon dioxide from the periodic table where we see we have 44 point oh one g for one mole of C 02. Now canceling out our mold of C 02, let's use the color red. We're now going to convert from grams to kilograms as our final answer here, where we would recall the conversion that we have 10 to the third power grams equal to one kg. This allows us to cancel out our gramps, leaving us with kilograms as our final unit. And this is going to yield the mass of carbon dioxide produced equal to 4.02 times 10 to the 11th power kg of CO2 produced with the yearly diesel consumption. And this would be our final answer to complete this example. I hope everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Watch next