Understanding Gas Reactions and Stoichiometry

Now that we've seen how we can mix gases together, let's see what happens if we let them react with each other after we let them mix. Okay, we'll start with the example we just did in the previous presentation. So if you haven't watched that yet, go back and watch the presentation about gas mixing. So we're going to use our oxygen and nitrogen monoxide apparatus. where the left side has the oxygen at 1.50 liters and 2.50 atmospheres and on the nitrogen monoxide side we'll a nitrogen monoxide side sorry we have six liters and 0.798 atmospheres um we're gonna open the valve oh and we're gonna do this one first like we did last time we're gonna do this first at a given temperature and later on we'll do this without stating what the temperature is and see if we can still solve the problem. So the temperature was 31.5 degrees celsius. We previously calculated the number of moles using this temperature and we got 0.150 moles of oxygen and 0.191 moles of nitrogen monoxide. When the valve is open we let the gases mix together. The new partial pressures are, here we go, 0.50 atmospheres and 0.637 atmospheres. And the number of moles, of course, don't change. We didn't add moles. We didn't take them away. Their pressures changed because the volumes changed. Temperature remains constant. And as we calculated in the last presentation, the partial pressure, the total pressure after mixing. but before they react with each other is 1.137 atmospheres. Okay, before reaction this is what we have. Now we have a chemical reaction between nitrogen monoxide and oxygen that is going to form nitrogen dioxide. First things well, so the question is after reaction What's the partial pressure and number of moles of nitrogen dioxide? This is a limiting reagent problem, so we're going to ask all the questions that we do when we do limiting reagent problems. What's the limiting reagent? What are the partial pressures of each gas after the reaction is complete, which means we need to know what the excess reagent is and how much of it is left. We need to know how much of the product. The NO2 is formed and then once we have that all of these are gases we calculate their pressures So that's how we're gonna do first things first balance this reaction, please. I hope you noticed it wasn't balanced So it is two moles of nitrogen monoxide Reacts with one mole of oxygen to form two moles of nitrogen Dioxide that's the balanced chemical reaction. Okay, so we have our before reaction What is our after reaction? So we're going to do our limiting reagent problem, just how we always do our limiting reagent problem. The way I like to do them, there are alternate ways to do them, but the way I like to do them is to just ask the question, if oxygen is the limiting reagent, the one point, if oxygen is the limiting reagent, the 0.150 moles of O2, is going to react at a stoichiometric ratio, right? So the mole ratio is two, sorry, two moles of NO2 formed for every one mole of O2 used. So I'm going to multiply by that fraction, two over one. And I can see that if oxygen is limiting, we will form 0.300 moles of NO2. Let's do the other one. If nitrogen monoxide is limiting, then the 0.191 moles of NO will react at a 2 moles of NO2 formed for every 2 moles. Here's the stoichiometric coefficient. Moles of NO used. So it's a 2 to 2 ratio, which means 0.191 moles of NO2 would be formed. We determine the limiting reagent by looking at which one of these two numbers is smaller. The smaller of the two is the bottom one. So we know that the limiting reagent is nitrogen monoxide. Okay, so nitrogen monoxide is limiting, so let's calculate the moles and partial pressures after this reaction is completed. We just calculated on the last slide that the moles of product that are going to be formed is 0.191 moles of nitrogen dioxide, so we know how much our product is. I'll put it down. here 0.191 moles. All of the nitrogen monoxide is used up, right? It's a limiting reagent. There's nothing left, so I know that there are no moles of nitrogen monoxide left, no pun intended. And then finally, for the oxygen we have to do our calculation for our excess reagent. Oxygen is excess So how much oxygen is consumed during this reaction? So we're going to do the 0.191 moles of NO reacts at a 1 to 2 ratio, right? A 1 to 2 ratio with the NO2, with the NO rather. And that means during the reaction, 0.0955 moles of O2 is consumed. We started with 0.150 moles. do the subtraction and we get 0.055 moles. This is not a significant figure, but I'm keeping it in it just in the calculations. But we would round this off in the answer. So that's how much is left of the oxygen. So I'll put that down here. We only get two significant figures in this answer because of the subtraction. Okay, so now we know how many moles are left. Now we know how many moles of each are left. All we have to do is use PV equals nRT to calculate the pressure of each gas. So PV equals nRT rearranged to P equals n times R times T over V, which remember is the combined volume of these two bulbs, and we get 0.637 atmospheres partial pressure for the NO2. This shouldn't be a surprising number, right, because all of the NO became NO2 in this reaction and the original pressure of the NO in this mixture was 0.637 after we allowed it to mix. So that's where our partial pressure of NO2 comes from. Next we're going to calculate the partial pressure of NO. Well, calculate. There's nothing left, right? Zero moles means there's zero atmospheres of partial pressure. And then finally, we'll use the 0.55 moles of oxygen that remain after the reaction to calculate the partial pressure of O2 as 0.18 moles. There we go. So now we have the partial pressure of each. We can put together all those partial pressures into a total pressure 0.637 plus 0 plus 0.18 and that's 0.82 atmospheres total pressure after the reaction. So that's how we do a gas phase reaction. It's We're just using PV equals nRT in this case to find our n, and once we have n all we have to do is do a regular stoichiometric calculation like we usually do. So that's how we solve these kinds of problems. Let's do that again. What if we weren't given the pressure and we can't find the number of moles like we did? in the last presentation where we were just told the pressure is constant. So what we're going to do is use the laws of partial pressures and we're going to use the stoichiometry of this reaction a little bit differently. What's the limiting range? So here's how we're going to read the stoichiometry of this problem. Because these gases are all in the same volume and they're all at the same temperature, what I can do is read the stoichiometry of this equation slightly differently. Instead of seeing this as two moles of NO2 react with one mole of O2 to form two moles of NO2, I'm going to say two atmospheres. of NO gas react with one atmosphere of O2 to form two atmospheres of NO. By doing this in terms of atmospheres, partial pressures rather than moles, because I can do that because I know there's a direct proportionality between the number of moles and the partial pressure in the same container at the same temperature. So if the 0.500 atmospheres of oxygen is the limiting reagent at a ratio where two atmospheres of NO2 will be formed for every one atmosphere of O2 that is reacted. The the pressure of O2, sorry the pressure of the product NO2 will be 1.00 atmospheres. If the NO is the limiting reagent, the 0.6 0.637 atmospheres of NO reacting at a two atmospheres to two atmospheres. NO2 to NO ratio will produce 0.637 atmospheres of NO2. Therefore, this is the smaller number. NO is the limiting reagent. The partial pressure of NO2 is just what we just calculated. 0.637. The partial pressure of the NO is zero because there's nothing left. And then we have to go back and calculate the remaining partial pressure of oxygen. The remaining partial pressure of oxygen we're going to calculate by saying that that original 0.637 atmospheres of nitrogen monoxide reacted with one atmosphere of for every two atmospheres of NO. So we used up 0.3185 atmospheres. We can't keep the significant figure. Atmospheres of O2. So the amount of oxygen remaining is the original 0.500 atmospheres minus that 0.3185 atmospheres, which leaves us with 0.18. 1815 atmospheres, which we have to round off to three significant figures this time, to 0.182 atmospheres of O2 remaining. So we get an extra significant figure, by the way, by doing the calculation this way that we didn't have in the last calculation. But we have the same numbers for the partial pressures, and we never needed to calculate the number of moles of oxygen or NO or the number of moles of NO2. for this reaction. If we don't know the temperature, we don't know how many there are, but we can still calculate the pressures in this reaction. Okay, so that's two ways to look at these gas law reactions. Let's do a couple more examples, or at least one more. Okay, let's do another gas law stoichiometry. In the previous example, it was all gases. but we can also use them to calculate reactions between phases. For instance, in alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide. According to the reaction, here is our glucose C6H12O6 solid that is going to react to form two moles of C2H 5OH, this is our ethanol molecules, plus two moles of CO2 gas. Carbon dioxide gas is also a product of fermentation. So we're going to ask a percent yield question, but we're going to use the gas laws in calculating our stoichiometry. So here's the question. If 5.97 grams of glucose are reacted and 1.44 liters of carbon dioxide Dioxide gas are collected at 293 Kelvin and at 0.984 atmospheres. What is the percent yield of the reaction above? Okay, so there are two steps to this problem, maybe three steps to this problem, but there's two hard steps and an easy step. We're going to use the stoichiometry of the reaction. to find the theoretical yield. So we're just going to say if this mass of glucose reacted completely then how many moles of CO2 would we have formed. Next we're going to use the gas laws in PV equals nRT and we'll use the ideal gas law approximation to calculate the actual yield how many moles of CO2 were collected in this experiment. Then we'll have those two different values to compare to each other. So we'll compare the actual value from part two to the theoretical value in part one. And we'll find the percent yield. So let's go through and do that. OK, step one, let's do just a regular, you know, problem like we did before. Our 7.97. grams of glucose where the molar mass of glucose is we're going to find 6 times 12.01 plus 12 times 1.008 plus 16 times 6 there it is equals 180.16 grams per mole so here's our molar mass for glucose so now we can do our calculation by looking at the mass of the glucose converting the mass using the molar mass one mole per 180.16 grams so this is going to give us the number of moles of glucose then we look at the stoichiometry of the reaction one mole of glucose yields two moles of carbon dioxide so times two moles of carbon dioxide to get that the theoretical yield if all of the glucose reacted to form carbon dioxide, we should get 0.0663 moles of carbon dioxide. Okay, so now we have our theoretical yield. Let's calculate our actual yield. Okay, we're going to use ideal gas law and the pressure, temperature, and volume that was given in the problem, which was 1.44 liters, 293 Kelvin, and point. 986 atmospheres. There are our values. We're going to rearrange PV equals nRT to solve for the number of moles of glucose. We're going to solve for the number of moles of carbon dioxide. So it's n equals P times V over R times T. So that's 0.984 atmospheres times 1.44 liters. divided by 0.08206 liters times atmospheres divided by moles divided by kelvin times 293 kelvin. The kelvins cancel, the atmospheres cancel, the liters cancel. We have 1 over inverse moles which is moles and the value is 0.0589. Okay, so now we have our actual yield and our theoretical yield. Ignore that. All we need to do is to find percent yield is actual yield divided by theoretical yield times 100%. Here are values of actual yield. Theoretical yield gives us 88.9% yield. So here we've used the gas laws to calculate our stoichiometric quantities and related them to a multi-phase problem. We had solids, we had liquids, we had gases, but n is just number of moles and number of moles can be used in any kind of stoichiometry problem that we find ourselves trying to solve. Okay so that's our gas stoichiometry laws.

So we're going to use our oxygen and nitrogen monoxide apparatus. where the left side has the oxygen at 1.50 liters and 2.50 atmospheres and on the nitrogen monoxide side we'll a nitrogen monoxide side sorry we have six liters and 0.798 atmospheres um we're gonna open the valve oh and we're gonna do this one first like we did last time we're gonna do this first at a given temperature and later on we'll do this without stating what the temperature is and see if we can still solve the problem. So the temperature was 31.5 degrees celsius.

We previously calculated the number of moles using this temperature and we got 0.150 moles of oxygen and 0.191 moles of nitrogen monoxide. When the valve is open we let the gases mix together. The new partial pressures are, here we go, 0.50 atmospheres and 0.637 atmospheres. And the number of moles, of course, don't change. We didn't add moles.

We didn't take them away. Their pressures changed because the volumes changed. Temperature remains constant. And as we calculated in the last presentation, the partial pressure, the total pressure after mixing.

but before they react with each other is 1.137 atmospheres. Okay, before reaction this is what we have. Now we have a chemical reaction between nitrogen monoxide and oxygen that is going to form nitrogen dioxide.

First things well, so the question is after reaction What's the partial pressure and number of moles of nitrogen dioxide? This is a limiting reagent problem, so we're going to ask all the questions that we do when we do limiting reagent problems. What's the limiting reagent?

What are the partial pressures of each gas after the reaction is complete, which means we need to know what the excess reagent is and how much of it is left. We need to know how much of the product. The NO2 is formed and then once we have that all of these are gases we calculate their pressures So that's how we're gonna do first things first balance this reaction, please. I hope you noticed it wasn't balanced So it is two moles of nitrogen monoxide Reacts with one mole of oxygen to form two moles of nitrogen Dioxide that's the balanced chemical reaction. Okay, so we have our before reaction What is our after reaction?

So we're going to do our limiting reagent problem, just how we always do our limiting reagent problem. The way I like to do them, there are alternate ways to do them, but the way I like to do them is to just ask the question, if oxygen is the limiting reagent, the one point, if oxygen is the limiting reagent, the 0.150 moles of O2, is going to react at a stoichiometric ratio, right? So the mole ratio is two, sorry, two moles of NO2 formed for every one mole of O2 used. So I'm going to multiply by that fraction, two over one.

And I can see that if oxygen is limiting, we will form 0.300 moles of NO2. Let's do the other one. If nitrogen monoxide is limiting, then the 0.191 moles of NO will react at a 2 moles of NO2 formed for every 2 moles. Here's the stoichiometric coefficient.

Moles of NO used. So it's a 2 to 2 ratio, which means 0.191 moles of NO2 would be formed. We determine the limiting reagent by looking at which one of these two numbers is smaller. The smaller of the two is the bottom one.

So we know that the limiting reagent is nitrogen monoxide. Okay, so nitrogen monoxide is limiting, so let's calculate the moles and partial pressures after this reaction is completed. We just calculated on the last slide that the moles of product that are going to be formed is 0.191 moles of nitrogen dioxide, so we know how much our product is. I'll put it down.

here 0.191 moles. All of the nitrogen monoxide is used up, right? It's a limiting reagent.

There's nothing left, so I know that there are no moles of nitrogen monoxide left, no pun intended. And then finally, for the oxygen we have to do our calculation for our excess reagent. Oxygen is excess So how much oxygen is consumed during this reaction? So we're going to do the 0.191 moles of NO reacts at a 1 to 2 ratio, right? A 1 to 2 ratio with the NO2, with the NO rather.

And that means during the reaction, 0.0955 moles of O2 is consumed. We started with 0.150 moles. do the subtraction and we get 0.055 moles.

This is not a significant figure, but I'm keeping it in it just in the calculations. But we would round this off in the answer. So that's how much is left of the oxygen. So I'll put that down here. We only get two significant figures in this answer because of the subtraction.

Okay, so now we know how many moles are left. Now we know how many moles of each are left. All we have to do is use PV equals nRT to calculate the pressure of each gas. So PV equals nRT rearranged to P equals n times R times T over V, which remember is the combined volume of these two bulbs, and we get 0.637 atmospheres partial pressure for the NO2. This shouldn't be a surprising number, right, because all of the NO became NO2 in this reaction and the original pressure of the NO in this mixture was 0.637 after we allowed it to mix.

So that's where our partial pressure of NO2 comes from. Next we're going to calculate the partial pressure of NO. Well, calculate.

There's nothing left, right? Zero moles means there's zero atmospheres of partial pressure. And then finally, we'll use the 0.55 moles of oxygen that remain after the reaction to calculate the partial pressure of O2 as 0.18 moles.

There we go. So now we have the partial pressure of each. We can put together all those partial pressures into a total pressure 0.637 plus 0 plus 0.18 and that's 0.82 atmospheres total pressure after the reaction.

So that's how we do a gas phase reaction. It's We're just using PV equals nRT in this case to find our n, and once we have n all we have to do is do a regular stoichiometric calculation like we usually do. So that's how we solve these kinds of problems.

Let's do that again. What if we weren't given the pressure and we can't find the number of moles like we did? in the last presentation where we were just told the pressure is constant.

So what we're going to do is use the laws of partial pressures and we're going to use the stoichiometry of this reaction a little bit differently. What's the limiting range? So here's how we're going to read the stoichiometry of this problem. Because these gases are all in the same volume and they're all at the same temperature, what I can do is read the stoichiometry of this equation slightly differently. Instead of seeing this as two moles of NO2 react with one mole of O2 to form two moles of NO2, I'm going to say two atmospheres.

of NO gas react with one atmosphere of O2 to form two atmospheres of NO. By doing this in terms of atmospheres, partial pressures rather than moles, because I can do that because I know there's a direct proportionality between the number of moles and the partial pressure in the same container at the same temperature. So if the 0.500 atmospheres of oxygen is the limiting reagent at a ratio where two atmospheres of NO2 will be formed for every one atmosphere of O2 that is reacted. The the pressure of O2, sorry the pressure of the product NO2 will be 1.00 atmospheres. If the NO is the limiting reagent, the 0.6 0.637 atmospheres of NO reacting at a two atmospheres to two atmospheres.

NO2 to NO ratio will produce 0.637 atmospheres of NO2. Therefore, this is the smaller number. NO is the limiting reagent. The partial pressure of NO2 is just what we just calculated.

0.637. The partial pressure of the NO is zero because there's nothing left. And then we have to go back and calculate the remaining partial pressure of oxygen.

The remaining partial pressure of oxygen we're going to calculate by saying that that original 0.637 atmospheres of nitrogen monoxide reacted with one atmosphere of for every two atmospheres of NO. So we used up 0.3185 atmospheres. We can't keep the significant figure. Atmospheres of O2. So the amount of oxygen remaining is the original 0.500 atmospheres minus that 0.3185 atmospheres, which leaves us with 0.18.

1815 atmospheres, which we have to round off to three significant figures this time, to 0.182 atmospheres of O2 remaining. So we get an extra significant figure, by the way, by doing the calculation this way that we didn't have in the last calculation. But we have the same numbers for the partial pressures, and we never needed to calculate the number of moles of oxygen or NO or the number of moles of NO2.

for this reaction. If we don't know the temperature, we don't know how many there are, but we can still calculate the pressures in this reaction. Okay, so that's two ways to look at these gas law reactions.

Let's do a couple more examples, or at least one more. Okay, let's do another gas law stoichiometry. In the previous example, it was all gases. but we can also use them to calculate reactions between phases.

For instance, in alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide. According to the reaction, here is our glucose C6H12O6 solid that is going to react to form two moles of C2H 5OH, this is our ethanol molecules, plus two moles of CO2 gas. Carbon dioxide gas is also a product of fermentation. So we're going to ask a percent yield question, but we're going to use the gas laws in calculating our stoichiometry. So here's the question.

If 5.97 grams of glucose are reacted and 1.44 liters of carbon dioxide Dioxide gas are collected at 293 Kelvin and at 0.984 atmospheres. What is the percent yield of the reaction above? Okay, so there are two steps to this problem, maybe three steps to this problem, but there's two hard steps and an easy step. We're going to use the stoichiometry of the reaction.

to find the theoretical yield. So we're just going to say if this mass of glucose reacted completely then how many moles of CO2 would we have formed. Next we're going to use the gas laws in PV equals nRT and we'll use the ideal gas law approximation to calculate the actual yield how many moles of CO2 were collected in this experiment. Then we'll have those two different values to compare to each other. So we'll compare the actual value from part two to the theoretical value in part one.

And we'll find the percent yield. So let's go through and do that. OK, step one, let's do just a regular, you know, problem like we did before. Our 7.97.

grams of glucose where the molar mass of glucose is we're going to find 6 times 12.01 plus 12 times 1.008 plus 16 times 6 there it is equals 180.16 grams per mole so here's our molar mass for glucose so now we can do our calculation by looking at the mass of the glucose converting the mass using the molar mass one mole per 180.16 grams so this is going to give us the number of moles of glucose then we look at the stoichiometry of the reaction one mole of glucose yields two moles of carbon dioxide so times two moles of carbon dioxide to get that the theoretical yield if all of the glucose reacted to form carbon dioxide, we should get 0.0663 moles of carbon dioxide. Okay, so now we have our theoretical yield. Let's calculate our actual yield.

Okay, we're going to use ideal gas law and the pressure, temperature, and volume that was given in the problem, which was 1.44 liters, 293 Kelvin, and point. 986 atmospheres. There are our values. We're going to rearrange PV equals nRT to solve for the number of moles of glucose.

We're going to solve for the number of moles of carbon dioxide. So it's n equals P times V over R times T. So that's 0.984 atmospheres times 1.44 liters. divided by 0.08206 liters times atmospheres divided by moles divided by kelvin times 293 kelvin.

The kelvins cancel, the atmospheres cancel, the liters cancel. We have 1 over inverse moles which is moles and the value is 0.0589. Okay, so now we have our actual yield and our theoretical yield. Ignore that.

All we need to do is to find percent yield is actual yield divided by theoretical yield times 100%. Here are values of actual yield. Theoretical yield gives us 88.9% yield.

So here we've used the gas laws to calculate our stoichiometric quantities and related them to a multi-phase problem. We had solids, we had liquids, we had gases, but n is just number of moles and number of moles can be used in any kind of stoichiometry problem that we find ourselves trying to solve. Okay so that's our gas stoichiometry laws.

Transcript for:Understanding Gas Reactions and Stoichiometry

Transcript for:
Understanding Gas Reactions and Stoichiometry