Limiting Reagent

in this series of videos, we're gonna still talk about Stoke geometry. But now we're gonna introduce the idea of the limiting re agent. Now, the limiting re agent sometimes called the limiting, reacted. It's the reacting that is completely consumed in a reaction and determines the maximum amount of product. Now we start talking about limiting react in a re agent when more than one reacted and our chemical equation is given. Ah, starting them out. Now we're going to say here that the limiting re agent helps us to determine the theoretical yield. This is the maximum amount of product that conform from a chemical reaction. It's also referred to as the 100% yield or the maximum yield now different from the limiting re agent is the excess re agent. Now, this is the reacting the remains after the completion of the chemical reaction. So you have a reactant and acts as a limiting re agent and then you can have another reactant acting as the excess, um, re agent. Now, in order to determine which reacting is which, you must work out the amounts of products each can make. And from there we'll have enough information to tell which is which. Now that we've gotten out the definitions, click on the next video and see how the Stoke geometric chart changes slightly when dealing with limiting re agents and excess region.

you might notice that this documentary chart is slightly different from the one we learned about in stock geometry in stock geometry were accustomed to getting on Lee. One given amount for reacted within our chemical reaction. Sometimes we even mentioned the other reacted. Now we're gonna have chemical equations with react INTs, and more than one of them will have a given amount. So what we're gonna have to do is we're gonna have to do start geometry for each reacted. So for reacting one, we're gonna take its amount and that amount might be in grams. So we'll have grams of given for it. Convert those grams have given two moles of given. We know that we have to do the jump to get two moles of unknown. And to do that, we have to use a coefficients from the balance equation. Those molds even know we can convert them into ions, atoms, formula units, molecules or Gramps. We get our answer for that amount of product or unknown. Then we'd have to do it again for reacting to go through the whole process and find out how much of our own known we have. So from those amounts that we decide we've calculated. We could determine which one of these reactions is the limiting re agent and which one is the excess re agent. We can also determine from the limiting re agent what the theoretical yield is is just double the work to help us find out how much product we're going to make. So click on to the next video. We'll let's take a look at example question where we put to practice this new idea of our story geometric chart.

in this example question. It says chromium three oxide reacts with hydrogen sulfide, which is H two s gas to form chromium three sulfide and water. All right, so here we have our chromium three oxide reacting with 33 hydrogen sulfides to produce chromium, three sulfide and three waters. Within this question, they asked what is the mass of chromium three sulfide formed when 14. g of chromium three oxide reacts with 12 80 g of hydrogen sulfide. Now, in this question, they're giving us mawr than 1 g of given. Okay, so we have 2 g of given and they're asked me to find grams of unknown because they're giving me multiple grams of given. I'm gonna have to do Stoke Yama tree for both of those amounts. So I'm gonna go through the whole process of the Stoke geometric chart with a 14.20 g and then again with a 12.80 g. Once we get our answers for the amount of chromium three sulfide produced, we'll talk about those two answers and which one is the final solution. So if we take a look here at these steps, we have step one we're going to convert the given quantities. So both those grams in tow moles of given now if any compound or compounds is said to be in excess, then just ignored. Yeah, once we have both of those moles have given, we do a multiple comparison to convert them into moles of unknown. And then Step three, we convert those moles of unknown finally into our desired units, which is grams of chromium three sulfide. All right, so let's set this up. We have 14.20 g of chromium three oxide. Let's just start out with this one. I'm gonna convert moles of it. I'm grands of it into moles. So one mole of chromium, three oxide on top. It's grams on the bottom here. I need to know how much does this way. So I have to look at the periodic table. So we have two chromium is and three oxygen's. According to the periodic table theater Tomic. Massive chromium is 51.996 g, and for oxygen and 16 Gramps, this comes out to 103 992 g and then this is 48 g. So together that's 1 51.992 g as the mass for chromium three oxide. So here, grams, cancel out Now I have moles of given we're gonna convert Those moles have given into moles of unknown So moles of our unknown is what we're asked to find which is chromium three sulfide Remember, at this point, this is where we utilize the jump So we look at the coefficients of the balance equation So it's for every one of this There is one of this so these moles cancel out And then finally we're gonna convert one mole of chromium three sulfide into grams of chromium three sulfide. We already calculated the mass for chromium three oxide and so a lot Let's do it for chromium three sulfide. Okay, so here we're gonna have three sull furs. Seoul for on the periodic table is 32 07 g, roughly so multiplying that by three gives me 96.21. So bring everything down here is gonna be zero 02 So that's 2.202 g of chromium three sulfide. So now these moles cancel out, and that equals when we work this out that's going to give me point 7 g of chromium three sulfide. So Corman three oxide says that it can make this much product. Let's see how much hydrogen sulfide says it can make. So we have 12.80 g of hydrogen sulfide. Okay, We have one mole of hydrogen sulfide on top and then grams of hydrogen sulfide on the bottom. When we add together the two hydrogen, each one is 1.8 g and the one sulfur, which is 32 07 g, Their combined mass is 34. g grams. Cancel out now I'm gonna go from moles of Given two moles of unknown. Yeah, And remember, when we make this jump, we look at the coefficients. So coming back up here for every three moles of hydrogen sulfide we have one of chromium three sulfide. So it's three and then one here these moles cancel out and then I already calculated the mass for chromium three sall fight. So just plug it back in. Okay, so when we do that, that's gonna give us 17 g of chromium three sulfide, so we have to totals here. How do we determine which one is the correct answer? Well, that's when we look at Step four. Compare the final amounts of the unknown to determine the theoretical yield. Remember, the theoretical yield represents your 100% yield. How much you could make maximum. We're going to say here that the smaller amount represents my living re agent, and the larger amount represents my Xs regent. So here are two totals. The smaller amount is 12.7 g. Those 12.7 g were produced by this chromium three oxide, So chromium three oxide represents my limiting re agent. The larger amount of 17 grams, 17 g was produced by hydrogen sulfide. Hydrogen sulfide represents my xs free agent. Remember, it is the limiting re agent that determines the theoretical yield. It is the one that determines how much product you could theoretically make if this reaction was working perfectly. So the limiting re agent is chromium three oxide, and so it's total amount of product it says it can make is our theoretical yield. So this would be the answer here. This is how much product we could potentially make if everything was working toe 100% efficiency, right? So just remember, if they're giving you multiple grams of given, moles have given any units of given, find out how much product each compound can make. The answer. The theoretical yield is always a smaller answer. That smaller answer is created by the limiting re agent. Okay, so in this question, 12.7 g of our product would be our final answer.