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Propane, $\mathrm{C}_{3} \mathrm{H}_{8},$ is a hydrocarbon that is commonly used as a fuel.(a) Write a balanced equation for the complete combustion of propane gas.(b) Calculate the volume of air at $25^{\circ} \mathrm{C}$ and 1.00 atmosphere that is needed to completely combletely combust 25.0 grams of propane. Assume that air is 21.0 percent $\mathrm{O}_{2}$ by volume. (Hint: we will see how to do this calculation in a later chapter on gases - for now use the information that 1.00 $\mathrm{L}$ of air at $25^{\circ} \mathrm{C}$ and 1.00 atm contains 0.275 $\mathrm{g}$ of $\mathrm{O}_{2}$ per liter.)(c) The heat of combustion of propane is $-2,219.2 \mathrm{kJ} / \mathrm{mol}$ . Calculate the heat of formation, $\Delta H_{\mathrm{f}}^{\circ}$ of propane given that $\Delta H_{\mathrm{f}}^{\circ}$ of $\mathrm{H}_{2} \mathrm{O}(l)=-285.8 \mathrm{kJ} / \mathrm{mol}$ and $\Delta H_{\mathrm{f}}^{\circ}$ of $\mathrm{CO}_{2}(g)=-393.5 \mathrm{kJ} / \mathrm{mol} .$(d) Assuming that all of the heased in burning 25.0 grams of propane is transferred to 4.00 kilograms of water, calculate the increase in temperature of the water.

aBalanced chemical equation for propane combustion is:$$\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$$b$$V_{\text {air }}=329.9 \mathrm{L}$$c$$\Delta_{f} H_{\mathrm{CaHs}, g}^{\circ}=-104.5 \frac{\mathrm{kJ}}{\mathrm{mol}}$$d$$\Delta T=75.2^{\circ} \mathrm{C}$$

Chemistry 101

Chapter 5

Thermochemistry

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Hello there. And welcome to problem number 85. Um, this problem is gonna be a little bit long because it does have four parts to it A, B, C and D s. So just bear with me and we will work through all four parts together. Started out for part A. They tell us we are starting with propane, and it is going through a combustion. It is combusting, and they would like us to write a balanced equation. So propane is C three h eight. Combustion means it is reacting with oxygen. He had the two products of a complete combustion are always going to be carbon dioxide and water. Okay, To balance this, I am going to need three carbons on the product side since I started with three. Hey, and I'm going to need eight total hydrogen. That gives me six plus four, which is 10 oxygen on the product side. So coming back over to the react inside, I'm going to add a five. And this is our balanced equation for the combustion. Okay, in part B, they want to know how many leaders or what the volume of air is that would be needed if we're trying to react 25 g of the propane and they tell us what percentage of air is oxygen, they give us a conversion there. So this is going to be a conversion problem. I'm gonna have to do some strike geometry, starting with my 25 g of C three h eight. I'm gonna go ahead and convert first two moles of C three h eight that I will use the mole ratio to give me two moles of oxygen. Um, Then I will use theme Mueller Mass of oxygen to figure out how many grams of oxygen are needed. And then I'll use the conversion that they gave me. That tells me how many grams of oxygen there are per leader of air. So that's my plan. Let's go ahead. But that into motion first thing I'm doing is converting from grams of propane, two moles of propane. I get a molar mass of 44.9 g of propane in every mole of propane. When I look at the periodic table and and the molar masses together for C three h eight, then using the mole ratio from the balanced equation, I see there's a coefficient of one in front of the C three h eight. I am interested in finding out information about the other reactant the oxygen and there's a five in the balanced equation in front of that. So that means for every one mole of C three a. J, I will need one mole of oxygen. Now I want to convert those moles of oxygen 2 g. So in every mole of oxygen, since it is 02 I need two times the molar mass of an oxygen atom. So two times 16 gives me 32 g of 02 And then finally, in the problem, they told me that zit there are 0.275 grams of oxygen in every leader of air. So that's just a direct conversion that I picked out of the problem. Alright. Looking at what cancels here, Grams will cancel, moles will cancel, moles of oxygen, will cancel, grams of oxygen will cancel and I am left with an answer. In terms of leaders of air, we are allowed three significant figures here, which means I have 330 with that zero being significant leaders of their and that would be my answer for part B in part C. They want us to calculate the heat of formation for this reaction. Remember the heat of formation? According to Hess's law, If we take the heat of If we take the NFL peas for the products and we subtract in this case the heat of combustion for from the reactant since we will get the heat of formation. So let's go ahead and do that. So I Delta h of formation I am going to take first the products. There are three moles of CO two and in the product and the problem. They tell us that there are negative 3 93 0.5 killer jewels for every mole. All right, so that is for the carbon dioxide. The other product is water. There are four moles of water. According to the balanced equation, take four moles of the H 20 times the NFL p value that they gave us in the problem. The heat of formation is negative to 85 0.8 killer jewels Permal. All right. This'll will give me the value for the products. Now I need to subtract the react. It's they give us the heat of combustion for the propane. There's only one mole of propane. According to the balanced equation, he and the heat of combustion for that is negative. 2219.2 killer jewels from all okay, Looking at the oxygen. Since it's a gaseous element, it's heat of formation is a zero. So I do not need to add in the oxygen. So this is my equation. How it's time to do some solving, grabbing my calculator. I'm gonna do this in a couple steps. I'm going to simplify. First of all, I'm going to calculate my products. I get negative 2300 and 23.7 killer jewels for the products. And then I am subtracting the reactant, which is negative. 2219 point to kill a jewels. Um, two negatives, of course. Make that a positive sign. So finishing this out? Yeah, I get my delta heat of formation to be negative. 104.5 killer jewels for every mole of C three h eight that combusts. And that would be my answer for part C. We are almost there. One more part. Alright. For part D in part d, we are trying to determine the change in temperature if we use 4 kg of water. So how much? How much will the temperature of that water change if we combust? 25 g of propane. Okay. Yeah. All right. Let's see here we are trying to figure out change in temperature. Looks like the equation I'm going to use is the amount of heat produced. He's going to be equal to the I'm sorry, that should be a little lower case C because I want specific heat here. So the specific heat of water times its mass in grams put him times the change in temperature. That's the equation I want to use because I can look up the specific heat of water. I have a massive water given, and I can calculate you based upon the information that we have in this problem. So let me go ahead and calculate que I need to calculate. Q is given. Our Q is going to be the heat energy produced by the combustion of this methane. That is, of course, given to me already in kill the jewels per mole. So I need to figure out how many moles of propane I have so that I could figure out how Maney killer jewels will be released. So starting with my 25 g, I need to calculate moles. So do that. I'm gonna take my 44 09 g, which is my Moeller massive C three h eight in every mole. Yeah. Then I'm going to take that heat of combustion we were given for propane. We were told that it is 2200 at 19.2 killer jewels. Permal. Since I'm ultimately determining how much energy is absorbed by the water, I'm going to use the positive value of that. Of course, the negative value tells you how much is given off. Now. I'm interested in how much is absorbed. I'm sorry. That is Permal got ahead of myself a little bit there since I was talking. All right, so that is per mole of the C three h eight. Finally, what I want to do is I want to express killer jewels in terms of jewels, because I know my specific heat is always given in jewels, so I'm just gonna convert. There's one killer Jewell is equal to 1000 jewels. This is going to give me an amount of heat in jewels and I get a pretty large number since I converted it to Jules. Alright, I get that many jewels. I am ready to plug all of my values into this equation and solve four Delta T. Let's go ahead and do that Cube. That's the amount of heat released or absorbed in this case. So that's the number of jewels I just finished calculating. That's gonna be equal to the specific heat of water looking that up in the table. It is 4.184 Jules program Degrees Celsius. Notice. This is Ingram's. So I want to express my 4 kg of water in terms of grams. So 4 kg doing a quick conversion here in 1 kg, there are 1000 grams finally time still to t so I can now solve. I could go ahead and solve this now for Delta T. And I find that the change in temperature three only unit that's gonna be left because Jules and Grams are going to cancel the only only unit remaining are gonna be degrees Celsius, and it is 75.2 degrees Celsius. And that is our answer for part D. Hey, thanks so much for sticking with me through the entire problem. Oh, I hope you found this helpful.

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