A gaseous fuel mixture contains 25.3% methane (CH₄), 38.2% ethane (C₂H₆), and the rest propane (C₃H₈) by volume. When the fuel mixture contained in a 1.55 L tank, stored at 755 mmHg and 298 K, undergoes complete combustion, how much heat is emitted? (Assume that the water produced by the combustion is in the gaseous state.)

Question

Accepted Answer

Welcome back everyone in this example, we're told that a certain fuel mixture is made up of the following gasses in percent by volume. Were given 30.6% of propane gas, 52.3% of butane gas and the rest is plantain gas. So we're told that the mixture is stored at 298 kelvin and 745 millimeters of mercury in a 2.5 liter zero liter gas tank. When the mixture is subjected to complete combustion, we need to calculate the total heat that will be admitted. So note that the water vapor is produced in the combustion reaction. So what we should recognize is that below were given the entropy of formation for each of our gasses below. And our first step is going to be recognizing that we know the pressure of the environment that are gasses are stored in which is given in the prompt here as 7 45 millimeters of mercury. But we need the individual pressures of our gasses. And so we want to start out part one of our solution by finding the partial pressure of propane. So that C three H eight. The partial pressure of butane given as C four H 10 and then our partial pressure of plantain C five H 12. So we want to recall upon our dalton's law of partial pressure. Now, recognize that in the prompt were given our gasses in% 100. And so what we would say is that for our propane we have a percent by volume of 30.6 which we can translate as 30.6 mm of Mercury. And this is going to be out of 100 of mercury. And so because we want to get our final unit for pressure, which we should recall is going to be in a t. M. We need our final unit to be a T. M. So we need to cancel out millimeters of mercury. And what that means is that we're going to use that info from the prompt being the pressure of the the environment that are gasses are stored in 45 millimeters of mercury as a conversion factor. So what we would say is that we should recall that for 1 80 M we have And equivalent of of mercury and plugging in the pressure of the environment that these gasses are in from the prompt. We would be able to cancel out our mm of mercury and the denominator. So we're just going to multiply by that 745 mm of mercury. And so canceling out our units, we can get rid of millimeters of mercury here in the numerator and here in the denominator as well as here in the denominator with this millimeters of mercury in the numerator. And this would leave us with a t m's as our final unit for partial pressure, which is what we want. And so in our calculators, this should give us a result equal to about 0.3000 A. T. M. So this is our partial pressure of our propane. And now we're going to go to the partial pressure of butane. So we also need to follow the same steps. So looking at its percent by volume given the prompt as 52.3, we would interpret that as 52.3 mm of mercury out of 100 of mercury. This is done going to be multiplied by our conversion factor. We recall that 1 80 M. And let's make this clear. So we recall that 1 80 M. Has an equivalent of 760 millimeters of mercury. Were able to cancel out our units of millimeters of mercury in the numerator and the denominator here and then we just have one more unit of millimeters of mercury to cancel out. So we will multiply by the pressure of the environment given in the prompt as 745 millimeters of mercury. So we can cancel that out now again we're left with a T. M's. And we should yield a result in our calculators equal to about 0.5127 ATMs as the partial pressure of butane. And lastly we have the partial pressure of plantain. So beginning with its percent by volume given in the prompt, we actually need to calculate that because we are not given in the prompt. So because we know that we have 52.3% of our butane. We're going to add that to our percent by volume of propane given us 30.6%. This total gives us 82.9% and this is for propane and butane. And so what that means is to find our percent by volume of plantain. We would assume that out of 100%,, We're going to subtract 82.9. So we'd say 100 -82.9 And this difference is going to give us our percent by volume of painting equal to just 17.1. So we can say 17.1% of our painting. And so we're going to utilize that and interpret that as a decimal 17. mm of mercury. This is out of 100 of mercury making up our mixture. And we're going to first multiply by canceling out the units millimeters of mercury in the numerator by recalling that 1 80 M is equivalent to 760 millimeters of mercury. We can cancel out the numerator of millimeters of mercury with the denominator here and to get rid of that millimeters of mercury in the denominator there. We're going to multiply. Sorry, multiply by our pressure of our environment that our mixture is stored in given in the prompt as 7 45 millimeters of mercury. So we can now cancel that out again. We're left with a t. M's. And this should yield us a partial pressure for plantain equal to a value of 0.1676 80 M. So now we have the partial pressure of each of our gas is we need to find the molds of each of our gas is in part two of our solution. So we called our ideal gas equation, which can give us the molds of each of our gasses. And we should recall that. That's going to be the pressure of our gas, multiplied by its volume, which is going to equal the molds of our gas times the gas constant R. Times the temperature in kelvin. Now we want to reorganize this to sulfur moles. So this is for our moles of each gas. So we would reorganize this and say that our molds of each gas should equal the pressure of that gas times its volume divided by gas constant R. And then multiplied by temperature in kelvin. So we'll begin by finding our moles of our first gas listed, which is our propane. So above we calculated its pressure or partial pressure as 0.3000 a. t.m. We want to multiply by the volume given to the prompt for our mixture as 2.50 L. And then in our denominator we're going to divide by R. Gas constant R. Which we should recall as a value of 0.8206 Leaders times a t M's divided by moles times kelvin. And we want to multiply by temperature given in the prompt as 298 Kelvin. So canceling out our units, we will get rid of our units of a T. M's in the numerator with a T. M. And the denominator here, we can also cancel out our units of leaders as well as our units of kelvin. Leaving us with our final unit in moles, which is what we want. And this is going to yield a result of 0.3067 moles of our gas propane. So moving on, we have our gas butane, so C four H 10. We want to find its moles. So above we did find its partial pressure which we calculated to be 0. 278 E. M. Sorry, that's a tm. There we multiplied by the volume of our mixture given in the prompt as 2. liters. And again we divide by R. Gas constant R which we recall is 0.8206 Leaders times A T M's divided by moles, times kelvin and then we will multiply by the temperature given in the prompt as 298 kelvin. Again canceling out our units, we get rid of a tm. We get rid of kelvin and we get rid of leaders were left with moles and we're going to get a result equal to a value 0. moles of our gas butane. So lastly we need our moles of our plantain. So C five H 12 and we would find that by taking its partial pressure which above we calculated at 0.1676 80 M We're going to multiply by the leader or volume of the mixture given in the prompt as 2.50 liters. We're going to divide by the gas constant R which we recall 0.8 to 06 leaders times a Tm. We divide by most times kelvin units and we will multiply this by the temp Given the prompt as to 98 kelvin again canceling out our units. We can get rid of a T. M. S, we can get rid of leaders and we can get rid of kelvin. We're left with moles. This gives us our molds of our propane or sorry pent ain't equal to a value of 0. moles of our plantain. Now we still need to move on in our solution because again, our final answer needs to be in units of kilograms because we need to give the total heat of our fuel mixture combustion. And so we want to find our entropy change for each of our gasses based on their combustion individually. So beginning with our propane we have C three H eight and this is for part three of our solution where we need to find the entropy the change in entropy for each of our gasses, we should recall that in a combustion reaction. Our gas is going to be reacting with oxygen gas and we always will produce products of carbon dioxide and water. Now we need to make sure that this is balanced. So we're going to place the coefficient of five in front of our oxygen gas, a coefficient of three in front of our carbon dioxide and a coefficient of four in front of our water. And now that we have our combustion for propane written out, we want to recall that to calculate the entropy change for our propane. We're going to take the difference between our entropy of formation of our products minus the entropy of formation of our reactant and this will give us our entropy of our reaction. So beginning with our propane, we can say that it's entropy is equal to first beginning with our products. So we'll use the color gray to express our products. We have our first product C. 02 where we have three moles of C. 02, multiplied by carbon dioxides entropy of formation given in the prompt as negative 393.5 kg jewels Permal, then added on to this. We have our second product where we have our water and just to be clear, they should have the labels gas, they're both gasses. So our product water is added on to this and we have formals multiplied by waters. Heat of formation given in the prompt as negative 241.8 kg joules per mole. Now moving on to our reactant side, we're going to subtract and say for our first reactant propane, which we just have one mole of. We have one mole, multiplied by propane xenophobia formation given in the prompt as negative one oh 3.85 kg joules per mole. And we're actually going to end off our calculation here because we only want to find the entropy change for propane, because our final answer needs to be the total heat of each of our gasses. So we're going to disregard our oxygen gas in this calculation and what we're going to get is our rental be equal to a value of negative 2043.85 kg joules. So moving on to our next gas, we have our butane C four H 10. We want to write out how it reacts in a combustion. So it would react with oxygen gas to form our products carbon dioxide gas and water. And to bounce us out, we would go ahead and place a coefficient of 13.5 over our oxygen gas on the react inside a coefficient of four in front of our carbon dioxide and a coefficient of five in front of our water for a balanced equation. So calculating the entropy change for butane, We would begin with our products 1st. So starting off with our four moles of our carbon dioxide and sorry, we can just say formals, we're going to multiply by carbon dioxide heat of formation equal to a value given in the prompt as negative 3 93. kg jewels Permal. This is then added to the formals or five moles rather of our second product water where we would multiply by water's heat of formation given in the prompt as negative 41.8 kg, joules per mole and now subtracting for our product side or our reactive side. Rather we have our one mole of our butane gas, multiplied by its heat of formation equal to a value given in the prompt as negative 125.7 kg jewels per mole. And just so it's visible, will scoot that over. So this will yield our entropy of our butane equal to a value of negative 2657.3 kg jewels. And just to be clear, all of our units of moles cancel out which is why we're left with kayla jewels as our final unit in both steps so far. So next we move on to our gas plantain, we have C five H 12. We recall that in a combustion it's reacting with oxygen to form carbon dioxide and water all gaseous products and react ints and to bounce us out. We're going to place a coefficient of eight in front of our gaseous oxygen a coefficient of five in front of our gaseous carbon dioxide and a coefficient of six in front of our gaseous water. And now to calculate the entropy change of our reaction for plantain. Beginning with our products, we have our five moles of our carbon dioxide multiplied by its entropy of formation given the prompt as negative 3 93.5 kg joules per mole. And then added on to this, we have our second product, six moles of gaseous water multiplied by its heat of formation given in the prompt as negative 41.8 kg joules per mole. And then subtracting for our reactant. Our plantain gas, we have just one mole of plantain gas multiplied by its heat of formation given in the prompt as negative 46.9 kg jewels per mole, canceling out all of our mole units were left with kayla jewels. And we're going to get a entropy of our reaction for planting gas equal to a value of negative 3271.4 kg joules. And so now we have all of the heats for each of our individual gasses or sorry entropy changes for each of our individual gasses and we can use these entropy values for each of our gasses to find our individual heats for each of our gasses in our mixture so we can use them as a conversion factor. And so going back to our molds of each gas beginning with propane. We have 0.3067 moles of our propane that we calculated above in part one of our solution. And we're going to multiply by the entropy change of propane in a combustion where we see that one mole of our propane gas has a entropy of negative 2043.85 kg joules. And this leaves us with killer goals as our final unit because we will be able to cancel out moles and this is what we want for our heat. And so this would give us a heat for propane equal to negative 62.68 kg joules. Moving onto our butane, we calculated above, we have 0.5241 moles of our butane. We will multiply by its entropy of in a combustion reaction as a conversion factor. We reset above that for one mole of butane, it has an entropy equal to negative 2657. kg jewels. So now we're able to cancel out moles were left with kayla jewels, which is what we want. And we're going to get a value equal to negative 1 39.27 kg joules. And lastly we have our plantain gas. So we calculated above that, we have 0.1713 moles of our plantain. We would multiply by its will be in a combustion reaction as a conversion factor, which above we stated for one mole of our plantain, we have an entropy of negative or negative 3271.4 kg jewels. We can cancel our units of moles were left with units of kilo jewels and we get a value equal to negative 56 point oh four kg jewels. And so our last step is to just add these totals up for each of our heats. And this is going to give us a total heat of our gaseous mixture equal to a value of negative 258 kg jewels. And this is going to be our final answer to complete this example as the total heat of our gaseous mixture. So I hope that everything I explained was clear. But if you have any questions, please leave them down below and I will see everyone in the next practice video

Verified Solution

Key Concepts

Combustion Reactions

Ideal Gas Law

Heat of Combustion