If 2.0 mol CH3CH2CH2COOH, 2.0 mol C4H10, and 2.0 mol C6H6 are completely combusted in oxygen, which one produces the largest number of moles of H2O?

Question

Accepted Answer

Hey everyone in this example, we're told that methane, ethane and propane are burned in excess oxygen in three separate separate reaction vessels were told that the mass burnt for each fuel was 3.45 g. And we need to determine which fuel will generate the largest amount of C. 02. So what this question is asking us to determine is our theoretical yield of carbon dioxide produced from our given re agents, methane, ethane and propane. We're told that these reactant are burned in excess oxygen. So we know that we have therefore a combustion reaction and we should recall that in a combustion reaction our products are going to be carbon dioxide and water. So we need to write out reactions for each of these given re agents. So first, beginning with methane, We would go ahead and recall that methane formula is CH four. It's reacting with oxygen to produce carbon dioxide and water in this combustion reaction. And we want to make sure that this is balanced. So we're going to place a coefficient of two in front of our water as well as the coefficient of two in front of our oxygen gas. To fully make sure that this is a balanced equation. Our next step is to find our moles of our methane and we're told that the mass burn of methane is 3.45 g. So we would say 3.45 g of methane and we want to go from grams of methane to moles of methane. So we would recall the molar mass from the periodic table for every atom and methane, which is equal to 16.04 g of methane for one mole of methane. And this is going to give us moles of methane equal 2.215 moles of methane. Now that we have moles of methane, we want to go ahead And find our grams of carbon dioxide produced from our .215 moles of our methane here. And so we want to go ahead and take note of our ratio from our reactant, methane to our reactant or our product, rather carbon dioxide and that comes from our balanced equation here. So we said, we have according to our coefficients, one mole of methane producing one mole of carbon dioxide, giving us a 1 to 1 molar ratio. Which is what we're going to use during our calculation. But first we want to go ahead and write out moles of methane To go into M.nolds of our product carbon dioxide. And then because we want the massive carbon dioxide as our final unit, we want to cancel out moles of carbon dioxide. So we're going to plug in our molar mass of carbon dioxide, which from our creator table we would get is equal to a value of 44.1 g of carbon dioxide for one mole of carbon dioxide. And then in the middle here we're going to plug in that molar ratio so that we can go from moles of methane to moles of our product carbon dioxide. And we said that that molar ratio is that for one mole of methane we produce one mole of carbon dioxide from our balanced equation. And so now we're able to cancel out our units And we're left with g or our mass of carbon dioxide. And this is going to be equal to a value of 9.46 g of carbon dioxide that is theoretically produced. And so now we want to do the same steps for our following two gasses. So, following the same process for ethane, we want to recall that our formula for ethane is C two H six. It's reacting with oxygen to produce in a combustion reaction carbon dioxide and water. And so we want to make sure that this is balanced. So we're going to add the following coefficients. We're going to add a two in front of our C2H6, a four in front of our CO2, a six in front of our water and then a seven in front of our oxygen gas. So that we have a fully balanced equation, which is always the first step Now that we have a fully bounced equation, we're going to take that mass for our fuel. So according to the problem, that's 3.45g of our ethane and we want to find moles of ethane. So we need to cancel out grams of ethane And we're going to refer to the periodic table for the molar mass of ethane equal to a value of 30.07 g for one mole of methane. And this is going to equal a value Of 0.115 moles of ethane. Our next step is to take note of our ratio of our re agent ethane To our product carbon dioxide. And according to our balanced equation here, we have two moles of ethane producing four moles of carbon dioxide. So we would have a 2-4 molar ratio. And so our next step is to go directly into that molar ratio as a conversion factor so that we can move from molds of ethane two moles of our product carbon dioxide. And so again, we said that we have two moles of methane producing four moles of carbon dioxide. And so the last step is to move from moles of carbon dioxide to find the mass grams of carbon dioxide produced. And so we would recall that our molar mass of carbon dioxide is still going to be from the periodic table 44 point oh one g for one mole of carbon dioxide. And now we can go ahead and cancel out our units, Leaving us with our mass of carbon dioxide as our final unit, which is going to give us a value equal to 10.1 g of carbon dioxide that is theoretically produced from ethane. And moving forward with our last given gas, we have propane. So this is our third fuel given, we should recall our formula for propane is C three H eight. It's reacting with oxygen gas and a combustion reaction to produce carbon dioxide and water. And we want to make sure that this is balanced. So we'll place a coefficient of four, sorry, a coefficient of four in front of our water. And that's an H. There. We're gonna place a coefficient of five in front of our oxygen gas And then a coefficient of three in front of our carbon dioxide. And this will ensure that everything is balanced in our given equation. And so according to the prompt, we still have that 3.45g of our propane used In our combustion. And so we're going to plug in the molar mass of propane in our denominator, which from the periodic table is 44.1 g of propane To get two moles of propane. So this is for one mole of propane according to our molar mass on the periodic table. And this is going to give us moles of propane equal to 0.0782 moles. And so now we can go and make note of our ratio between our propane and our product carbon dioxide and this is going to come from our balanced equation. So we see from our coefficients we have one mole of propane producing three moles of carbon dioxide, giving us a 123 molar ratio. And so we would say that we have in our denominator, one mole of our propane. And to get two moles of carbon dioxide, we know that that one mole of propane from our ratio produces three moles of carbon dioxide. And then as before we have our plugged in molar mass from the periodic table of carbon dioxide To cancel out moles of carbon dioxide. And we said that is 44.01 g per one mole of carbon dioxide. We are allowed to cancel out our units now Leaving us with our massive carbon dioxide produced, giving us a value equal to 10.3 g of CO produced theoretically from our propane. And sorry, just going to scoot this over so that all of its visible. And so to answer this question, we need to figure out which fuel generated the largest amount of carbon dioxide theoretically. And so what we would say is that our 9. g of carbon dioxide is less than the 10. g of carbon dioxide that was produced, which is even less than 10.3 g of carbon dioxide that was produced from our fuel propane. And so to complete this example, our final answer is going to be that propane was the fuel that produced the most, or the highest mass of carbon dioxide at 10.3 g of carbon dioxide. So this will complete this example as our final answer. I hope that 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.

If 2.0 mol CH3CH2CH2COOH, 2.0 mol C4H10, and 2.0 mol C6H6 are completely combusted in oxygen, which one produces the largest number of moles of H2O?

Verified Solution

Key Concepts

Combustion Reaction

Stoichiometry

Molar Mass and Molecular Formula