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# Stoichiometry

Introduction to stoichiometry. Using the balanced reaction to find molar ratios. . Created by Sal Khan.

Video transcript

We know what a chemical equation is and we've learned how to balance it. Now, we're ready to learn about stoichiometry. And this is an ultra fancy word that often makes people think it's difficult. But it's really just the study or the calculation of the relationships between the different molecules in a reaction. This is the actual definition that Wikipedia gives, stoichiometry is the calculation of quantitative, or measurable, relationships of the reactants and the products. And you're going to see in chemistry, sometimes people use the word reagents. For most of our purposes you can use the word reagents and reactants interchangeably. They're both the reactants in a reaction. The reagents are sometimes for special types of reactions where you want to throw a reagent in and see if something happens. And see if your belief about that substance is true or things like that. But for our purposes a reagent and reactant is the same thing. So it's a relationship between the reactants and the products in a balanced chemical equation. So if we're given an unbalanced one, we know how to get to the balanced point. A balanced chemical equation. So let's do some stoichiometry. Just so we get practice balancing equations, I'm always going to start with unbalanced equations. Let's say we have iron three oxide. Two iron atoms with three oxygen atoms. Plus aluminum, Al. And it yields Al2 O3 plus iron. So remember when we're doing stoichiometry, first of all we want to deal with balanced equations. A lot of stoichiometry problems will give you a balanced equation. But I think it's good practice to actually balance the equations ourselves. So let's try to balance this one. We have two iron atoms here in this iron three oxide. How many iron atoms do we have on the right hand side? We only have one. So let's multiply this by 2 right here. All right, oxygen, we have three on this side. We have three oxygens on that side. That looks good. Aluminum, on the left hand side we only have one aluminum atom. On the right hand side we have two aluminum atoms. So we have to put a 2 here. And we have balanced this equation. So now we're ready to do some stoichiometry. So the stoichiometry essentially ... If I give you... There's not just one type of stoichiometry problem, but they're all along the lines of, if I give you x grams of this how many grams of aluminum do I need to make this reaction happen? Or if I give you y grams of this molecule and z grams of this molecule which one's going to run out first? That's all stoichiometry. And we'll actually do those exact two types of problems in this video. So let's say that we were given 85 grams of the iron three oxide. So 85 grams. So my question to you is how many grams of aluminum do we need? How many grams of aluminum? Well you look at the equation, you immediately see the mole ratio. So for every mole of this, so for every one atom we use of iron three oxide we need two aluminums. So what we need to do is figure out how many moles of this molecule there are in 85 grams. And then we need to have twice as many moles of aluminum. Because for every mole of the iron three oxide, we have two moles of aluminum. And we're just looking at the coefficients, we're just looking at the numbers. One molecule of iron three oxide combines with two molecule of aluminum to make this reaction happen. So lets first figure out how many moles 85 grams are. So what's the atomic mass or the mass number of this entire molecule? Let me do it down here. So we have two irons and three oxygens. So let me go down and figure out the atomic masses of iron and oxygen. So iron is right here, 55.85. I think it's fair enough to round to 56. Let's say we're dealing with the version of iron, the isotope of iron, that has 30 neutrons. So it has an atomic mass number of 56. So iron has 56 atomic mass number. And then oxygen, we already know, is 16. Iron was 56. This mass is going to be 2 times 56 plus 3 times 16. We can do that in our heads. But this isn't a math video, so I'll get the calculator out. Is that right? That's 48 plus 112, right, 160. So one molecule of iron three oxide is going to be 160 atomic mass units. So one mole or 6.02 times 10 to the 23 molecules of iron oxide is going to have a mass of 160 grams. So in our reaction we said we're starting off with 85 grams of iron oxide. How many moles is that? Well 85 grams of iron three oxide is equal to 85 over 160 moles. So that's equal to, 85 divided by 160 equals 0.53125. Equals 0.53 moles. So everything we've done so far in this green and light blue, we figured out how many moles 85 grams of iron three oxide is. And we figured out it's 0.53 moles. Because a full mole would have been 160 grams. But we only have 85. So it's .53 moles. And we know from this balanced equation, that for every mole of iron three oxide we have, we need to have two moles of aluminum. So if we have 0.53 moles of the iron molecule, iron three oxide, then we're going to need twice as many aluminum. So we're going to need 1.06 moles of aluminum. I just took 0.53 times 2. Because the ratio is 1:2. For every molecule of this, we need two molecules of that. So for every mole of this, we need two moles of this. If we have 0.53 moles, you multiply that by 2, and you have 1.06 moles of aluminum. All right, so we just have to figure out how many grams is a mole of aluminum and then multiply that times 1.06 and we're done. So aluminum, or aluminium as some of our friends across the pond might say. Aluminium, actually I enjoy that more. Aluminium has the atomic weight or the weighted average is 26.98. But let's just say that the aluminium that we're dealing with has a mass of 27 atomic mass units. So one aluminum is 27 atomic mass units. So one mole of aluminium is going to be 27 grams. Or 6.02 times 10 to 23 aluminium atoms is going to be 27 grams. So if we need 1.06 moles, how many is that going to be? So 1.06 moles of aluminium is equal to 1.06 times 27 grams. And what is that? What is that? Equals 28.62. So we need 28.62 grams of aluminium, I won't write the whole thing there, in order to essentially use up our 85 grams of the iron three oxide. And if we had more than 28.62 grams of aluminium, then they'll be left over after this reaction happens. Assuming we keep mixing it nicely and the whole reaction happens all the way. And we'll talk more about that in the future. And in that situation where we have more than 28.63 grams of aluminium, then this molecule will be the limiting reagent. Because we had more than enough of this, so this is what's going to limit the amount of this process from happening. If we have less than 28.63 grams of, I'll start saying aluminum, then the aluminum will be the limiting reagent, because then we wouldn't be able to use all the 85 grams of our iron molecule, or our iron three oxide molecule. Anyway, I don't want to confuse you in the end with that limiting reagents. In the next video, we'll do a whole problem devoted to limiting reagents.