 In this video I want to go over solution stoichiometry. So we're going to talk about how we use these dilution units that we've learned about in the solutions chapter in calculations. So first of all as a review I just want to go over real quickly what molarity is. Molarity denoted by a uppercase m is a term that we use to describe the number of moles of our solute in one liter of solution. So molarity is equal to moles of solute over liter of solution. This is molarity and this is what you'll come into contact with during solution stoichiometry problems. So let's look at a quick example. Here we have a 5.5 molar hydrochloric acid solution. So what does that mean? Well if it says it's 5.5 molar then that means we have 5.5 moles of HCl in one liter of solution. And we can also write this in a different way. We can say that there are 5.5 moles in one liter of solution. Now you'll notice that this is an equality. An equality is something that we use in conversions. Remember in equality we can get two conversion factors from every equality. Conversion factor is simply writing an equality in the form of a fraction. So our two conversion factors from this equality would be 5.5 moles over one liter or one liter over 5.5 moles. Now we'll use these conversion factors to answer a question similar to this. If we're asked to convert 10 liters of a 5.5 molar HCl solution to moles, let's say I wanted to make 10 liters of a hydrochloric acid solution that had a concentration of 5.5 molar. How many moles of HCl do I need in order to do this? In order to set up this conversion we find out. Look at the problem and determine what it gives us. So it tells us that we have 10 liters of solution. This is what we're wanting to make or this is what we're wanting to convert from. We're going to multiply that 10 liters by a conversion factor that will allow us to convert to moles of HCl. Now since we know that our solution is 5.5 molar, we're going to pick one of these two conversion factors in order to complete this conversion. So the conversion factor that we're going to pick is this first one. It's going to go right here. Now why did we pick that conversion factor? We picked it so that our units cancel out. You've got solution, leaders of solution here, leaders of solution here. Those variables will cancel out. The only unit that we're left with is moles of HCl and that's what we're trying to convert to over here. So you take your 10 multiplied by 5.5 and then divide by one and we find that we need 55 moles of HCl in order to make 10 liters of a 5.5 molar HCl solution. So this is just the introduction to using molarity in these stoichiometric problems. Let's look at a little bit more difficult problem to solve. So we have 0.025 liters of a potassium chloride solution that has a molarity of 0.500 molar. And we're going to add this solution to a silver nitrate solution according to this reaction. And we want to know what mass of silver chloride will we form by adding this potassium chloride solution. What you'll notice here from our balanced equation is that we have potassium chloride and aqueous soluble salt. We're adding that to silver nitrate, also an aqueous soluble salt, to form potassium nitrate, aqueous soluble salt, and silver chloride. But silver chloride is insoluble so it forms an insoluble salt that will precipitate out of solution. What I want to know is if I add this 0.025 liters of potassium chloride to my silver nitrate, how much silver chloride can I form from this reaction? So the first step is we need to determine a pathway to answering this question. We're given the volume of potassium chloride in our question. It tells us that we're adding 0.025 liters of this potassium chloride solution. It also gives us the molarity of the potassium chloride solution. So we can convert from volume of potassium chloride to moles of potassium chloride. In order to do this conversion, the conversion factor that we're going to use is similar to the one that we used here. It's the conversion factor from that equality which tells us how many moles of solute are in a liter of solution. So in order to answer this conversion, we use the molarity of the potassium chloride that it gives us in the problem. However, this does not answer the problem. We have to do yet another conversion. So from moles of potassium chloride, we can then convert to moles of silver nitrate, or I'm sorry, silver chloride. This is what we're eventually wanting to get to. If we know the number of moles of potassium chloride, we can convert to the number of moles of silver chloride using our mole to mole ratio. Like I said, in order to do this conversion, we're going to use the mole to mole ratio which comes from our balanced equation. This gives us the moles of silver chloride that we can produce from this reaction, but yet once again, that does not solve our problem because the problem asks what mass of silver chloride we're going to need or we're going to form in this reaction. So we have a third conversion. We can go from moles of silver chloride to mass, mass of silver chloride. Any time we're going from moles of a compound to mass of that same compound, we're going to use the molar mass of that compound for this conversion. The molar mass comes directly from the periodic table. In order to determine the molar mass of silver chloride, you're going to get the mass, the molar mass, or the atomic mass of silver from the periodic table and add that to the atomic mass of chlorine also from the periodic table. So let's start this conversion here. We've got 0.025 liters of our potassium chloride solution. Our first conversion factor, we're going to use molarity of that potassium chloride solution in order to solve for the number of moles of KCl. So the molarity of the potassium chloride from our original problem is 0.500 molar. That also means it's 0.500 moles of potassium chloride per liter, moles per liter. You'll see that our liter here cancels out with liter and the value that we're left with is moles of potassium chloride. From this reaction, we find that a half molar solution of potassium chloride with a volume of 0.025 liters gives us 0.0125 moles of potassium chloride. That's our first conversion. Our second conversion will start with the answer to our last conversion. And for this conversion, we need to convert from moles of potassium chloride to moles of silver chloride. We've already said in order to do this, we're going to use the mole-to-mol ratio coming from our balanced chemical equation. Here's our balanced chemical equation, the coefficients that go in front of each reactant and product is what tells us our mole-to-mol ratio. You will see here that one mole of potassium chloride will produce one mole of silver chloride. So that is our mole-to-mol ratio of potassium chloride to silver chloride, is a 1-to-1 ratio. This is going to be our second conversion factor here. One mole of potassium chloride can form one mole of silver chloride. So in order to solve for this equation, we take our 0.0125 moles of potassium chloride, we multiply it by one and then divide by one. And our units that cancel out are going to be our moles of potassium chloride. So this time we are solving for moles of silver chloride. When we solve this conversion, we find it's equal to 0.0125 moles of silver chloride. And then finally, our last conversion, I'm going to start with the answer to our previous 0.0125 moles of silver chloride. We need to convert moles of silver chloride to grams of silver chloride. Like I said, you are going to use the molar mass for this conversion to go from moles of a compound to mass of the same compound. The molar mass of silver chloride is 143.4 grams of silver chloride per mole of silver chloride. This is going to make up our third conversion factor. 143.4 grams of silver chloride per one mole. Look at our units that cancel out once again, moles of silver chloride cancel out with moles of silver chloride. The unit we are left with is the unit that we are looking for which is our mass or our gram value of silver chloride. We take our 0.0125 moles. Let's apply that by 143.4, then divide by 1, and we find that we can make 1.79 grams of silver chloride when we add 0.025 liters of potassium chloride that has a molarity of 0.500 molar. Hopefully this helps with your calculation questions coming from the solutions chapter dealing with stoichiometry of solutions.