Calculating Mass of Excess Reactants

In this video, we're going to talk about how to find the mass of the excess reactant that remains after the reaction is complete. Now, I've broken up the process into three parts. The first thing you need to do is identify the limiting reactant. Next, you need to find out how many grams of excess reactant is consumed in a reaction. Now, once you complete steps 1 and 2, then you can find the answer to part C, or step 3. and that is finding the mass of the excess reactant after the reaction is complete. You want to find out how much is left over. So let's begin. The first thing we need to do is write a reaction and balance the equation. Aluminum is placed in a solution with sulfuric acid. Sulfuric acid is H2SO4, and it's going to produce hydrogen gas and aluminum sulfate. Now, what is the chemical formula of aluminum sulfate? aluminum sulfate. Aluminum has a positive 3 charge, sulfate has a minus 2 charge. So therefore the correct formula is Al2SO43. Now we need to balance the chemical equation. Notice that we have three sulfate ions on the right side and one SO4 unit on the left. So we've got to put a 3 in front of H2SO4. So we have three sulfate units. Now we have two aluminum atoms on the right, so we'll need a 2 in front of Al. Now we have six hydrogen atoms on the left, so we've got to put a 3 in front of H2 to have six hydrogen atoms on both sides. Now the equation is balanced. Let's put a 1 in front of this one. So, 80 grams of aluminum is placed in a solution that contains 40 grams of sulfuric acid. If we're given the grams of each reactant, how can we find the limonene reactant? To do so, take... Take the mass and divide it by the molar mass. The molar mass of aluminum is 26.98. When you divide the mass by the molar mass, this will give you the number of moles. So we have 2. Now let's find the molar mass of H2SO4. So there are two hydrogen atoms, one sulfur atom, and four oxygen atoms. So that's 2 times 1.008. Sulfur is 32.07 and oxygen has an atomic mass of 16. So 4 times 16, that's 64 plus 32.05. seven plus two times 1.008 so the molar mass of H2SO4 is 98.086 so let's take 40 and divided by that number so this will give us point 4078 so now we need to do clearly we can see that the limy reactants going to be sulfuric acid it's a lot less than aluminum So what we need to do now is divide by the coefficient. If we divide 0.4 by 3, we're going to get a smaller number than aluminum. So, sulfuric acid has the lowest mole per coefficient ratio. So therefore, H2SO4 in this example is the limiting reactant. So now that we finished part A, let's move on to part B. So I'm going to write this as the limnireactant, which means aluminum is the excess reactant. Now how many grams of the excess reactant is consumed in a reaction? To do that, you need to do a gram-to-gram conversion. Convert the grams of the limiting reactant to the grams of the excess reactant. This will tell you how many grams of the excess reactant is consumed in a reaction. So let's start with the 40 grams of sulfuric acid. And let's convert it to moles. Now the molar mass of sulfuric acid, we said it was 98.096. Now we need to change the substance. So let's use the molar ratio between sulfuric acid and aluminum. It's a 3 to 2 ratio. So for every 3 moles of sulfuric acid that reacts, 2 moles of aluminum reacts with it. Now let's convert moles of aluminum to grams using the molar mass of Al which is 26.98. So I got 7.33 grams of aluminum. So this is the answer to part B. That is the amount of excess reactant that is actually consumed in this reaction. So that's how much aluminum is used up in this reaction. So, if 7.33 grams of aluminum is consumed in a reaction, what is the mass of the excess reactant that remains after the reaction is complete? Let's make some space. So we started with 80 grams of reactant. During the course of the reaction, 7.33 grams of aluminum was consumed. So the amount that's left over, or the amount that remains, is going to be the difference of 80. and 7.33. So if we subtract those two numbers, this will give us the answer to part C. It's 72.67 grams of aluminum. So that's how much excess reactant remains after the reaction. That's how much is left over. For practice, go ahead and try this one. See if you understand this concept. Feel free to pause the video and see if you can get the right answer. So 50 grams of aluminum is placed with 70 grams of elemental sulfur, S8, to produce aluminum sulfide. Identify the aluminum reactant. So we have aluminum reactant with sulfur, and it produces aluminum sulfide. Now what is the chemical formula of aluminum sulfide? Aluminum has a positive 3 charge. sulfur has a negative 2 charge so it's going to be Al2S3. Now we need to balance the chemical equation. The least common multiple between 8 and 3, or just 8 times 3, is 24. So we need 24 sulfur atoms on both sides. Let's put a 3 in front of S8 and an 8 in front of AL2S3. So we have 24 atoms of sulfur on the left and on the right side. Now, 8 times 2 is 16, so we need a 16 on the left. So now the number of aluminum and sulfur atoms are the same on both sides. Now we're mixing 50 grams of aluminum with 70 grams of sulfur. Which one is going to be the limiting reactant? Is it the one that has less grams or more grams? Is it aluminum or sulfur? What do you think? Well first, let's find the moles of each substance. So let's take the grams and divide it by the molar mass. The molar mass of aluminum is 26.98. And for sulfur, it's going to be 8 times the atomic mass of sulfur. which is 32.07 so the molar mass of the s8 molecule is 256.56 50 divided by 26.98 that's 1.853 moles and 70 divided by 256.56 That's 0.2728 moles. So which one do you think is the limiting reactor at this point? Is it aluminum or sulfur? We have a lot more moles of aluminum. But let's see. Now that we have the quantity of aluminum and sulfur, we'll need to divide it by the coefficient. The coefficient in front of aluminum is 16. In front of sulfur, it's 3. So 0.2728 divided by 3 is 0.0909. 1.853 divided by 16, that's 0.1158. So in this case, sulfur is the limiting reactant. It's close to aluminum, but it has the lowest mole per coefficient ratio. So even though we have a greater amount of mass of sulfur, it doesn't mean it was a limiting reactant. You can't look at the grams and determine the limiting reactant. You need to find the lowest mole per coefficient ratio. Now let's move on to the next part. How many grams of excess reactant is consumed in this reaction? So we know this is the limiting reactant and this is the excess reactant. To answer part B, take the grams of the limiting reactant and convert it to the grams of the excess reactant. If you do it this way, the grams of excess reactant that you get should be less than how much you started with. If you try to do it the other way, you probably get more than the other reactant. So just always start with the limiting reactant. So we're going to start with 70 grams of elemental sulfur. And let's convert it to moles. So we know that the molar mass is 256.56. And now let's change the substance from the limited reactant to the excess reactant using the molar ratio. So for every 3 moles of sulfur that reacts, 16 moles of aluminum will react with it. Now let's convert moles of aluminum to grams using the atomic mass of aluminum, which is 26.98. So it's going to be 70 divided by 256.56, and then multiply that by 16, and divide it by 3, and then multiply that result by 26.98. So this will give you 39.26 grams of aluminum. So this is the amount of aluminum that is consumed in a reaction. That's the answer for part B. Now let's move on to the last part. What is the mass of the excess reactant that remains after the reaction is complete? So the excess reactant is aluminum. We started with 50 grams of aluminum, and during the course of the reaction, 39.26 grams of aluminum was consumed in reaction. So the amount that's left over is the difference between 50 and 39.26. So 10.74 grams of excess reactant remains in this reaction. So hopefully these two problems gave you a rubric to follow. It will help you to understand how to find the amount of excess reactant that remains. Hopefully you understand everything that we went over.

Next, you need to find out how many grams of excess reactant is consumed in a reaction. Now, once you complete steps 1 and 2, then you can find the answer to part C, or step 3. and that is finding the mass of the excess reactant after the reaction is complete. You want to find out how much is left over.

So let's begin. The first thing we need to do is write a reaction and balance the equation. Aluminum is placed in a solution with sulfuric acid.

Sulfuric acid is H2SO4, and it's going to produce hydrogen gas and aluminum sulfate. Now, what is the chemical formula of aluminum sulfate? aluminum sulfate. Aluminum has a positive 3 charge, sulfate has a minus 2 charge.

So therefore the correct formula is Al2SO43. Now we need to balance the chemical equation. Notice that we have three sulfate ions on the right side and one SO4 unit on the left. So we've got to put a 3 in front of H2SO4.

So we have three sulfate units. Now we have two aluminum atoms on the right, so we'll need a 2 in front of Al. Now we have six hydrogen atoms on the left, so we've got to put a 3 in front of H2 to have six hydrogen atoms on both sides.

Now the equation is balanced. Let's put a 1 in front of this one. So, 80 grams of aluminum is placed in a solution that contains 40 grams of sulfuric acid. If we're given the grams of each reactant, how can we find the limonene reactant? To do so, take...

Take the mass and divide it by the molar mass. The molar mass of aluminum is 26.98. When you divide the mass by the molar mass, this will give you the number of moles. So we have 2. Now let's find the molar mass of H2SO4. So there are two hydrogen atoms, one sulfur atom, and four oxygen atoms.

So that's 2 times 1.008. Sulfur is 32.07 and oxygen has an atomic mass of 16. So 4 times 16, that's 64 plus 32.05. seven plus two times 1.008 so the molar mass of H2SO4 is 98.086 so let's take 40 and divided by that number so this will give us point 4078 so now we need to do clearly we can see that the limy reactants going to be sulfuric acid it's a lot less than aluminum So what we need to do now is divide by the coefficient. If we divide 0.4 by 3, we're going to get a smaller number than aluminum. So, sulfuric acid has the lowest mole per coefficient ratio.

So therefore, H2SO4 in this example is the limiting reactant. So now that we finished part A, let's move on to part B. So I'm going to write this as the limnireactant, which means aluminum is the excess reactant.

Now how many grams of the excess reactant is consumed in a reaction? To do that, you need to do a gram-to-gram conversion. Convert the grams of the limiting reactant to the grams of the excess reactant. This will tell you how many grams of the excess reactant is consumed in a reaction.

So let's start with the 40 grams of sulfuric acid. And let's convert it to moles. Now the molar mass of sulfuric acid, we said it was 98.096.

Now we need to change the substance. So let's use the molar ratio between sulfuric acid and aluminum. It's a 3 to 2 ratio.

So for every 3 moles of sulfuric acid that reacts, 2 moles of aluminum reacts with it. Now let's convert moles of aluminum to grams using the molar mass of Al which is 26.98. So I got 7.33 grams of aluminum.

So this is the answer to part B. That is the amount of excess reactant that is actually consumed in this reaction. So that's how much aluminum is used up in this reaction. So, if 7.33 grams of aluminum is consumed in a reaction, what is the mass of the excess reactant that remains after the reaction is complete?

Let's make some space. So we started with 80 grams of reactant. During the course of the reaction, 7.33 grams of aluminum was consumed.

So the amount that's left over, or the amount that remains, is going to be the difference of 80. and 7.33. So if we subtract those two numbers, this will give us the answer to part C. It's 72.67 grams of aluminum.

So that's how much excess reactant remains after the reaction. That's how much is left over. For practice, go ahead and try this one. See if you understand this concept. Feel free to pause the video and see if you can get the right answer.

So 50 grams of aluminum is placed with 70 grams of elemental sulfur, S8, to produce aluminum sulfide. Identify the aluminum reactant. So we have aluminum reactant with sulfur, and it produces aluminum sulfide.

Now what is the chemical formula of aluminum sulfide? Aluminum has a positive 3 charge. sulfur has a negative 2 charge so it's going to be Al2S3. Now we need to balance the chemical equation.

The least common multiple between 8 and 3, or just 8 times 3, is 24. So we need 24 sulfur atoms on both sides. Let's put a 3 in front of S8 and an 8 in front of AL2S3. So we have 24 atoms of sulfur on the left and on the right side.

Now, 8 times 2 is 16, so we need a 16 on the left. So now the number of aluminum and sulfur atoms are the same on both sides. Now we're mixing 50 grams of aluminum with 70 grams of sulfur. Which one is going to be the limiting reactant?

Is it the one that has less grams or more grams? Is it aluminum or sulfur? What do you think? Well first, let's find the moles of each substance.

So let's take the grams and divide it by the molar mass. The molar mass of aluminum is 26.98. And for sulfur, it's going to be 8 times the atomic mass of sulfur. which is 32.07 so the molar mass of the s8 molecule is 256.56 50 divided by 26.98 that's 1.853 moles and 70 divided by 256.56 That's 0.2728 moles.

So which one do you think is the limiting reactor at this point? Is it aluminum or sulfur? We have a lot more moles of aluminum. But let's see.

Now that we have the quantity of aluminum and sulfur, we'll need to divide it by the coefficient. The coefficient in front of aluminum is 16. In front of sulfur, it's 3. So 0.2728 divided by 3 is 0.0909. 1.853 divided by 16, that's 0.1158.

So in this case, sulfur is the limiting reactant. It's close to aluminum, but it has the lowest mole per coefficient ratio. So even though we have a greater amount of mass of sulfur, it doesn't mean it was a limiting reactant.

You can't look at the grams and determine the limiting reactant. You need to find the lowest mole per coefficient ratio. Now let's move on to the next part.

How many grams of excess reactant is consumed in this reaction? So we know this is the limiting reactant and this is the excess reactant. To answer part B, take the grams of the limiting reactant and convert it to the grams of the excess reactant.

If you do it this way, the grams of excess reactant that you get should be less than how much you started with. If you try to do it the other way, you probably get more than the other reactant. So just always start with the limiting reactant.

So we're going to start with 70 grams of elemental sulfur. And let's convert it to moles. So we know that the molar mass is 256.56.

And now let's change the substance from the limited reactant to the excess reactant using the molar ratio. So for every 3 moles of sulfur that reacts, 16 moles of aluminum will react with it. Now let's convert moles of aluminum to grams using the atomic mass of aluminum, which is 26.98.

So it's going to be 70 divided by 256.56, and then multiply that by 16, and divide it by 3, and then multiply that result by 26.98. So this will give you 39.26 grams of aluminum. So this is the amount of aluminum that is consumed in a reaction.

That's the answer for part B. Now let's move on to the last part. What is the mass of the excess reactant that remains after the reaction is complete? So the excess reactant is aluminum.

We started with 50 grams of aluminum, and during the course of the reaction, 39.26 grams of aluminum was consumed in reaction. So the amount that's left over is the difference between 50 and 39.26. So 10.74 grams of excess reactant remains in this reaction. So hopefully these two problems gave you a rubric to follow. It will help you to understand how to find the amount of excess reactant that remains.

Hopefully you understand everything that we went over.

Transcript for:Calculating Mass of Excess Reactants

Transcript for:
Calculating Mass of Excess Reactants