Many home barbeques are fueled with propane gas (C₃H₈). What mass of carbon dioxide (in kg) is produced upon the complete combustion of 37.8 L of propane (approximate contents of one 5-gallon tank)? Assume that the density of the liquid propane in the tank is 0.621 g/mL. (Hint: Begin by writing a balanced equation for the combustion reaction.)

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Hi everyone here, we have a question telling us that lighters commonly used Beauty and gas C four H 10 for fuel. And our goal is to calculate the mass of carbon dioxide in grams produced upon the complete combustion of 56.0 mL of butane fuel fluid, approximate contents of a refill bottle. And we're going to assume that the density of the fluid is 0.573 g per milliliter. So first we're going to calculate the mass of beating and we're gonna do that by taking our amount 56 mL Times our density, which is 0. g per mil leader. And our ml will cancel out and that gives us 32. g a butane. Next, we need to write our balanced combustion reaction. So we have butane and because it's a combustion reaction, we're going to add oxygen and it is a di atomic adam. So there's two of them and that is going to form carbon dioxide plus water as a byproduct. So to balance this, let's ride out what we have on each side. So for carbon We have eight or we have four for hydrogen, we have 10 and oxygen, we have to and on our product side we have one carbon, two hydrogen and three oxygen. So we're gonna start this by putting a four in front of our carbon dioxide and that's going to change our carbon to four And our oxygen to nine. And next we're going to change our Hydrogen to five. So we have 10 hydrogen And we have 13 oxygen And as we can see with oxygen on our react inside there is a two. So we can't get 13. So we're going to have to multiply everything by two. So we're gonna change this to eight and this to 10, which gives us eight carbons, 20 hydrogen And oxygen's. And now we can balance the react inside by putting a two in front of our butane and a 13 in front of our oxygen. So we have eight carbon, 20 hydrogen And 26 oxygen. Now we have a balanced combustion reaction. So now we want to calculate the mass of carbon dioxide produced. And to do that, we first have to calculate the molar mass of butane and carbon dioxide. So our molar mass of butane, we have carbon, Which is 12.01 times four Equals 48.04. We have hydrogen which is 1. times 10 Which is 10.1 For a total of 58.14 g per mole. And all of these numbers are taken off the periodic table. And now we're going to do the same for carbon dioxide. So we have carbon which is 12. And we have oxygen which is 16 and we have two of them, Which is For a total of 44.01 g per mole. And now we're going to do our dimensional analysis. So we have .09 g a butane times one mole a butane over the molar mass of butane. And that's going to convert it into moles. So over 58. g of butane times eight moles of carbon dioxide over two moles of butane. And that is from our balanced combustion reaction times .01g of carbon dioxide. That's its molar mass Over one mole of carbon dioxide. And our grams of butane are going to cancel out. Our moles of butane are going to cancel out and our moles of carbon dioxide are going to cancel out, giving us 97.2 g of carbon dioxide. And that is our final answer. Thank you for watching. Bye.

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Key Concepts

Combustion Reaction

Stoichiometry

Density and Volume Conversion