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limiting reactant are accidents that control the number of moles that you can obtain for your product. And the smallest number of moles of your reactant will dictate the moles that you will obtain when getting the quantitative information from your chemical reaction. So this will be best illustrated by a couple of examples. Eso. Let's say you have a reaction where you have ammonium chloride reacting with calcium nitrate to get ammonium nitrate and calcium chlorate. So this reaction technically isn't a reaction just because all of these are unequally a solution, and we'll talk a bit more about these types of reactions in the next topic. But for the sake of this example, let's pretend this is an actual reaction. Um, and so let's say you have again ammonium chloride reacting with calcium nitrate, and you are given that there are 5 g of each material, so you need to figure out what is the limiting regent. So to do this, we need to first calculate the molar Mass and then from the Mueller Mass. We can use the grams that were given to obtain the number of moles, and then, from there we can compare the number of moles so that he reacted with e lower amount of moles will be the limiting reactant. So first, let's figure out the Mueller mass of ammonium chlorate. So ammonium chloride eyes made up of nitrogen for Hodgins and one chlorine. And so we can use the periodic table to figure out the values of each atom on, Then figure out the Mueller mess. So for nitrogen, this would be 14.1 and then you have four atoms of hydrogen. So then you need to multiply the mass of one hydrogen by four and finally the massive chloride. And when you do out this calculation, you should get a Mueller Mass of 53.49 grams Permal. So for this reaction were given that we have 5 g of both reactions and so we can use this information to calculate the number of moles given the mass of each reacted. So here we use dimensional analysis to figure out the number of bulls. So then we can divide the mass by the molar mass of, in this case, ammonium chloride, which will equals 0.9 3 47 months and now to figure out the limiting reactant. We also need to figure out the Mueller mass and number of moles or calcium metric. So doing the same thing from before calcium is 40.8 according to the periodic table. And in this case, you want to make sure that your, um adding the number of atoms correctly. And so here we see that nitrate is a poly atomic ion and there are two units of it. Eso the total number of message in is to and the total number of oxygen is so you wanna make sure that you include this in your calculation. So here I have two times 14, 14, um, corresponding to the nitrogen. And then you want to multiply six by the atomic mass of oxygen. And once you do this calculation, you should get a Moeller Mass of 164.1 g per mole, proceeding with the same method as before. We can divide 5 g by the molar mass of calcium nitrate, and so we end up with a mammal of 0.30 47 miles. And now we have determined the number of moles for ammonium chloride as well as calcium nitrate. And so we can now compare these values well, right, Thumbs down here. And this is for again ammonium chloride. And this is for calcium nitrate. And so if we compare these moles, we'll see that we have a greater number of moles for ammonium chloride in comparison to calcium nitrate. And so that means that the ammonium chlorate is in excess and the moles in calcium metric is limiting. And so that means that the limiting reacted for this reaction is calcium nitrate. So let's do another example of this. So let's say now you have a, um, acid reacting to a base, which we'll talk a bit more about later in the next topic. Eso Let's say you have something like sulfuric acid reacting with sodium hydroxide to yield sodium. So fate and water. Yeah, that's quickly, uh, the phases as well. Okay, Okay. So let's say in this case, we are given that 2.1 grands, uh, sulfuric acid reacts with one point 7 g of sodium hydroxide, and we need to figure out the eliminating reacted for this reaction. So you're going to use the same method as before, so we're going to calculate the molar masses of each reactant, find the number of moles and then from the number of moles, determine which one is the limiting reacted. And so again, we're going to find the molar mass. So there are two hydrogen atoms, one sulfur atom and for auction atoms. So if we figure out the Mueller mess, yeah, will be 98.9 grandes Permal. And now, using the mass that were given, we can calculate the number of moles for this reacted. So we'll divide the Mass by the molar mass of sulfuric acid, which will give us 0.2141 holds, and now we're going to use the same method as before. So we'll calculate the molar mass of sodium hydroxide, so that would be one sodium waas, one oxygen and one hydrogen, which will give us a Mueller Mass. 40.0 grams Permal. And then we're going to use the the mess that we're given and divide this by the molar mass, which will give us zero point zoo 4 to 50 moles. So the different thing about this reaction is that, um, so in the previous reaction, we had a 1 to 1 ratio of the reactant. But in this problem we have a ratio of 1 to 2. So basically, for everyone will of sulfuric acid, you need to react. Um, this reactant with two moles of sodium hydroxide. Consider this problem. We actually need to normalize, um, de moles of sodium hydroxide because we have more of it reacting with thes sulfuric acid and so we can divide this by two. And so this gives us a final mole 0.0 2125 boss. So now if we write this out again so moles of sulfuric acid is 0.2141 and the moles of sodium hydroxide is zero point 0 to 1 to five. We can see that there is actually a greater number of moles, uh, sulfuric acid, then sodium hydroxide and so sodium hydroxide is the limiting reactant in this equation. And for my last example, let's say that we are looking at the combustion of sugar. So again, for this reaction, we have sugar reacting with oxygen Thio create carbon dioxide and water and again we'll include our freezes has shown here and let's say we are given 1 g sugar, so 1 g of sugar and 0.1 g of oxygen gas. And we're trying to figure out the limiting reactant for this reaction. And so here, Well, again using the same method as before eso. In our previous problems, we've determined that the Miller Mass of sugar is 342.3 g per mole. And so we can use this to figure out the number of balls and so were given 1 g. So we can use this and divide this by the molar mass of sugar to obtain a final mole number of 0.0 29 to 1 and doing the same for oxygen auction has a molar mass. 32 grams Permal, and so we can use the grams that were given and divide this by the molar mass of oxygen to obtain 0.0 31 five malls of oxygen. So now, like the previous reaction, we need to normalize this because for every one mole of sugar, we need to react 12 moles of oxygen and so we need to divide us by 12. And what we end up with is two point 64 times 10 to the negative fourth malls. And now, to make the comparison easier to dio, we can convert this using scientific invitation. So this would be 2.9 to 1 times 10 something a third on this makes sense because, um, if we to buy this buy 1000, the decimal place moves over three times. And so this is correct. And so now we can compare the values that we obtained. So this is your and for oxygen gas, we get 2.64 times 10 to the negative, right? Eso. Now, if we compare this, we can see that sugar has a credible then oxygen gas and so we can conclude that auction gas is eliminating reacted. And so whenever you're trying to figure out the limiting reacted for an equation, you want to determine the number of grams that you are given so you can use the Mueller Mass to obtain the new moles. And after normalizing the moles, you will compare the bulls that you've obtained. Um, between the reactions and from there determined the smallest number of moles, which will be your answer

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