CHEMISTRY 5124-ZAMBIAN SYLLABUS- MOLE CONCEPT- LIMITING REAGENTS 1

Welcome to today's lesson. So we are looking at science paper two, which is chemistry following the ECZ standard. So our topic for today's lesson is stoichiometry and more concepts. So there are a lot of concepts that are in stoichiometry and more concepts and we'll be looking at these concepts one by one. So our subtopic in today's lesson is limiting reagents. So the sample question that we are going to use has been extracted from 2016 science paper two, which is question four B, question B4, which means section B. Okay, so now the first question simply says, what is meant by limiting reactant? Now, as we are trying to explain what limiting reactants are, we first of all have to give ourselves a quick refresh of the chemical reactions. So for every reaction, A plus B is going to produce C plus D. So we have what we call the reactants and we have the products. Now the reactants are the things that are reacting to produce what is supposed to be produced. So under this topic, limiting reagents, we are more interested in these ones here. So by definition is a limiting reagent is simply a reagent that is in short supply by more ratio. Okay, so what we mean by more ratio is, for example, when you write a chemical equation, it always has to be balanced. So use the coefficients to balance. So let's say this one has a coefficient A, this one B, this one C, and this one D. So to express this equation in terms of its more ratios, we are simply going to say number of moles of A divided by A is equal to number of moles of B divided by B, which is equal to number of moles of C divided by C, which is equal to number of moles of D divided by D. Okay, so that is just how it is done. Now I know it might look ambiguous right now. So we are going to try and deal with the actual calculations from the equation that we have on the board. Okay, so the first question has already been answered, which is A will go to B. So B, the question reads, 2.4 grams of magnesium reacts with 0.3 moles of hydrochloric acid. Write the balanced chemical equation for the reaction. Now there are two things that are reacting here. We have magnesium and we have hydrochloric acid. Now this one is a metal and this one is an acid. So to write a proper equation, you need to know the general rule, which is that every time a metal reacts with a dilute acid, we are going to have a salt and hydrogen gas. Okay, so now in this case, to answer the question specifically, we know the metal being given is magnesium. So we are going to say magnesium, which is a solid, is reacting with hydrochloric acid, which is aqueous. This is going to produce magnesium chloride, which is also aqueous and the hydrogen as a gas. Then to balance this equation, we just have to check. Magnesium here, we have one. There we have one. Hydrogen, we have one. Hydrogen, we have two. Chlorine there, two. Chlorine there, one. So to balance this equation, we are simply going to put a two. Okay, and your question B1 has been answered. This is the balanced chemical equation. So having balanced this one, the second question reads, determine the limiting reagent. Now to determine the limiting reagent, always be in the habit of writing a balanced chemical equation, then you express it into its more ratios. So this one will be number of moles of Mg over one, because there's a coefficient of one, is equals to number of moles of HCl over two, because of which is equals to number of moles of MgCl2 over one, which is equals to number of moles of hydrogen gas over one. So we have expressed it. Now, because we are determining the limiting reagent, our interest is on the reactor side. So now what we are going to do here is we are going to say AraMg is equals to number of moles of magnesium over one, this part here. So we try to find now what this one we are testing, whether this one is the one which is limiting or not. So we are going to say number of moles is equals to mass over molar mass. So this one here is equals to, so the mass has been given from the question, which is 2.4 grams. So 2.4 grams over the molar mass we get from the periodic table, which is 24 grams per mole. So when we divide this one and this one, our answer is going to be 0.1 moles. So we are found for this one. So we also test the second one. So we are going to say AraHCl is equals to number of moles of HCl over two. Now in the question, this one has been given to us, they are saying reacts with 0.3 moles. So we're just going to put 0.3, then divide this one by two, and our answer is equals to 0.15 moles. So we are found this one. So now to find the limiting reagent here, you just have to find the one which has got lesser number of moles compared to the other. Because remember, definition is that limiting reagent is the one that is in short supply by mole ratio. So we check which one is big between this one and this one. So we can conclude and say, since AraMg is less than AraHCl, then Mg is the limiting reagent. And this is how you find your final answer. Okay, let's see what the last question says. So to find the last question, let's create a subspace. Okay, so to find the last question here, they are simply saying calculate the mass in excess for the substance which is in excess. Now we've already figured out to say the one which is in excess is the HCl, since this one is limiting. So now to find the mass of this one, we know for sure to say number of moles is equals to mass over molar mass. Now what we are dealing with here is the excess. Now the question is how do we find the excess moles? So number of moles, the excess moles here will simply be equal to, you subtract the two that you have found at first. So which is 0.15 moles minus 0.1 moles. And your answer will be 0.05 moles. So these are the excessing moles. Now that you have the excess moles, you can now use it into this formula to say number of moles is equals to mass over molar mass. So over one, you cross multiply this one, you're going to find that mass is equals to number of moles. So for the excess multiplied by the molar mass. So this one will be 0.05 moles multiplied by now the molar mass of HCl. So the molar mass of HCl, we know HCl, this is 35.5. And this one is one. So when we add those, we are going to get 36.5 grams per mole. So we multiply this one by 36.5 grams per mole. Then this is chemistry, the beauty of it is you can even use a calculator. So when you use a calculator there, you find that your value is 1.825 grams because the more and the more will be cancelled. So you have determined the mass in excess for the substance which is in excess. Okay, now we have learned some concepts from this video. We can do a practice question from Science Paper 2, 2018 question A7, meaning it's in section A. So do that practice question, then you can give us the feedback. Thank you so much for your attention. Till next time.