The complete combustion of octane, C8H18, a component of
gasoline...

Question

The complete combustion of octane, C8H18, a component of
gasoline, proceeds as follows:
2 C8H181l2 + 25 O21g2¡16 CO21g2 + 18 H2O1g2
(c) Octane has a density of 0.692 g>mL at 20 C. How many
grams of O2 are required to burn 15.0 gal of C8H18 (the
capacity of an average fuel tank)?

Accepted Answer

Alright folks, so we have a balanced reaction below here that shows the combustion of propane. Right? So we have propane combusting with oxygen producing C. 02 and water. So if we're told that we have 132 point grams of C three H eight of propane, that's completely combusted. We need to calculate the mass in grams of oxygen needed for this process. So we want Massive oxygen given the mass of propane. Okay, so let's go ahead and start. So we already told that this is a balanced reaction. So we don't need to check. But you can go ahead and check anyways, it is balanced and we are ready to start. So 132.1 grams of propane. That's what we're going to start with. Because that's the only thing that we have given to us that we know. And we're gonna go ahead and convert this into moles so that we can do a multiple comparison between propane and oxygen to find uh the mass of 02. Alright, so in one mole of propane, how many grams do we have? That is just the molar mass of propane. Let's go ahead and find that on the side. So we have carbon and hydrogen For the carbon, we have three of them. Right? And each of them weigh 12. So that gives us 36 Gramps. And then for the hydrogen is we have eight of them. They weigh 1.01 each. So that is 8.08 g. And then in total, this will be 44.08 grams Permal. So that's the molar mass of propane. So we're going to put that on the bottom. So 44.08 grams of C three H eight In one mole of it. Okay, so g cancel out. And now we have moles. Now we can do a multiple comparison, right? So we can see that there is only one mole of propane. We're going to put that on the bottom so that it can cancel out. So for every one mole of propane we're using five moles of 02. So moles of O to go on the top because that's what we want. Right? So now that we have moles of 02, we can very easily convert that into grams right into mass and grams. So again, we just need the molar mass of 02. So in one mole of 02 and moles of O. To go on the bottom. So that we can cancel that out. How many grams are there? Well as oxygen, it's 16 g. Right? There's two of them. 16 times two is 32. So there are 32 g of 02 in one mole of 02. So moles cancel out and now we have grams of 02. And that will be our answer. So let's go ahead and divide and multiply everything correctly here. And it should give you 479.5 g of co two. Alright, so that's gonna be our answer for the mass and grams of CO. Two needed for this combustion reaction. Thank you so much for watching.

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