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Another stoichiometry example in a solution

Video transcript
Let's do another stoichiometry problem where the reaction is occurring inside of a solution. So we have a metallic zinc and it reacts with aqueous hydrochloric acid. We see that right there, the metallic zinc. And then the hydrochloric acid is dissolved in water, it's an aqueous solution. The solvent is water. Makes these products right there. And they say, what volume of 0.25 molar-- so they're giving us the concentration of the solution. What volume of 0.25 molar hydrochloric acid in milliliters-- so we've got to remember that, they want their answer in milliliters-- is required to convert 11.8 grams of zinc completely to products? So the way we can do this we'll figure out. We know that we're starting off with 11.8 grams right there. And we just have to figure out well, if we have 11.8 grams of zinc, how many moles of zinc that is. And then, assuming this is a balanced equation-- and I'm going to verify that right after I finish this sentence-- but assuming that this is a balanced equation, for every mole of zinc we're going to need two moles of hydrochloric acid. And then we can figure out given how many moles of hydrochloric acid we need, then how much of the solution do we need based on its concentration? So let's do that. It's going to, I think, be a lot simpler when we actually do it. But I just want to give you the action plan before we break into it. But before we do anything, let's verify that this is really a balanced equation. So you have 1 zinc on the left-hand side, 1 zinc on the right-hand side, 2 hydrogens on the left, 2 hydrogens on the right, 2 chlorines on the left, 2 chlorines on the right. So it looks balanced. Let's do the problem. So the first thing we want to do-- we're starting off with 11.8 grams of zinc, they give us that information. Now we're going to figure out how many moles of zinc that is. We're going to want to convert this to moles. So it's going to be moles of zinc over grams of zinc moles per gram. And if you look it up on a periodic table, zinc has an atomic weight of 65.38, so its molar mass is 65.38. So it's 1 mole for every 65.38 grams, or its 1/65, roughly, of a mole per gram. Actually, you know what? I'm just going to save our calculator for the end. So the grams of zinc cancels that with the grams of zinc, and we're left with 11.8 divided by 65.38 moles of zinc. Now we're trying to figure out how much of this solution of hydrochloric acid is required. So the next thing we want to figure out is how many moles of hydrochloric acid are required. And we see that for every mole of zinc we need 2 moles of the hydrochloric acid. Let me write it this way. For every 1 mole of zinc-- we'll put that in the denominator. I'll switch colors just arbitrarily here-- we need 2 moles of hydrochloric acid. That is required. And so then we can multiply the moles of zinc. We'll cancel out the moles of zinc. That's why I wrote 2/1 instead of 1/2. Two moles of hydrochloric acid for every mole of zinc. And what does this give us? This gives us-- I'm not even going to do the math. It's 11.8 times 2 over 65.38. And then we're left with moles of hydrochloric acid required-- moles of hydrochloric acid-- HCl-- required. So we haven't solved the problem yet. We want to figure out what volume of the solution we need. We know the actual number of molecules we need, but we need to know how much solution will contain that number of molecules. So we want to express this in liters. Well, eventually we're going to get to milliliters, but let's just go straight to liters first because molarity is always expressed in liters. So let's multiply this. We want the moles of HCl to cancel out. So let's put that in the denominator, moles of hydrochloric acid. And we want liters of hydrochloric acid, let me put it this way, liters of 2.50 molar hydrochloric acid solution. When anyone puts this big M there, that's molarity. You assume they're talking about a solution because they're giving you the concentration of the solute in the total solution. So how many moles are there per liter? Well this concentration is telling us. Molarity is moles per liter. So there are 2.50 moles for every 1 liter of our solution. That's what that tells us. This is a 2.5 molar solution. Its molarity is 2.5. There are 2.52 moles for every liter. And I wrote the moles in the denominator because I wanted to cancel out here. So it's 1/2.5. So let's cancel those two guys out. That cancels out with that right there. And then we are left with 11.8 times 2 in the numerator. And in the denominator, we have 65.38 times 2-- let me write the times not as a-- I'll write it that way-- times 2.50. And all we're left with in our units is liters of 2.50 molar HCl solution-- and then in that green-- required. We're almost done. Then we can get break out the calculator and get the exact answer. They want the answer in milliliters. So far we have the answer in liters. I mean, there's going to be some number here, I don't know what it is. But it's that number of liters of the 2.5 molar hydrochloric acid solution is required. Let me just copy and paste this. Let's convert this to milliliters because that's what they want. So that's what we had from the previous line. And to convert this to milliliters-- we've done this drill before-- we want the liters to cancel out. I'll just do it like this. And we want our answer in milliliters. And how many milliliters are there for a liter? Well there are 1,000 milliliters per 1 liter. And all of these steps-- the way I'm showing it to you, I'm making the dimensions cancel out. But hopefully, they're making some intuitive sense as well, that we divide by 2.5 at this stage because we have more than 1 mole per liter. So however many moles we require, we're going to have fewer liters required. So we're going to be dividing by a number that's greater than 1. If that doesn't make sense to you, just stick to the units. But at every one of these steps I really want you to actually conceptualize the numbers as well. But let's go back to the problem. So we're converting it to milliliters. So our final solution we're left with, we have the liters. Cancelling out the liters, and so we're left with, in the numerator, 11.8 times 2 times 1,000 over 65.38 times 2.50. And in the numerator we're left with milliliters of 2.5 molar hydrochloric acid. I'll just write it milliliters of 2.50 molar hydrochloric acid. I don't want this 2.50 molar HCl to confuse you. These aren't numbers in the computation, this is a label to say what type of solution. We could have just written over here liters of the solution in question. In case these numbers confuse you, we could have written liters of the solution. And then this would be milliliters of the solution. I don't want that 2.5 to confuse you. That's just the name of the solution, or its concentration. That's how we're identifying it. But now we've solved the problem. We just need to calculate this number, so let's break out the calculator. And you get 11.8 times 2 times 1,000 divided by 65.38. And we're also going to divide by 2.5. It's in the denominator, it's 144. And actually our lowest-- let's see, a couple of numbers here only have three significant digits, so we should really only stick to three significant digits, so it's 144 milliliters. So this right here is 144 milliliters of, we could say the solution, or of the 2.5 molar hydrochloric acid solution. And we're done. Hopefully you found that fun.