Hello, students, let us continue our discussion on Mole and Equivalent Concept.
Now in this module, first we will take an example on percentage purity and then we will discuss what is the percentage yield of a reaction. Let us take the example first.
An impure 6 grams of NaCl is dissolved in water and then treated with excess of silver nitrate. The mass of AgCl precipitate is found to be 14.35 grams. What is the percentage purity of NaCl? Now, in these cases first of all we will see what is the reaction. It is a double displacementry. It is a double displacementry reaction NaCl plus silver nitrate gives you AgCl plus NaNO3 and now, we will first develop the strategy here. What should be our strategy? Means any AgCl is precipitated, it will precipitate with the actual NaCl. So, using the mass AgCl precipitate we will first calculate the mass of NaCl that is present and then using this information we find the percentage purity. We have actual and theoretical it is given purity 6 gram, we can calculate. So, we write down the reaction and then first we do stoichiometric analysis. So, 1 mole NaCl gives you 1 mole AgCl. So, using the molar mass data, now they have given the atomic rates. Silver is 108, sodium is 23 and chlorine is 35.5. So, from here we can calculate the molar mass of NaCl as 58.5 and that of AgCl as 143.5. Now, from here converting moles into mass, we can write down 58.5 grams of NaCl is producing 143.5 grams AgCl. Now, we will do reverse analysis. 143.5 grams of AgCl is going to be produced by 58.5 grams of NaCl. So, 1 gram of AgCl is going to be produced by 58.5 divided by 143.5 grams of NaCl. Now, we are getting 14.35 grams of AgCl. So, that is going to produce, rather that is going to be produced by 58.5 by 143.5 into 14.35 grams of NaCl, which comes out as 5.85 grams of NaCl. This was actual amount. Theoretically how much is given, 6 grams. So, percentage purity is simply what 5.85 divided by 6 times 100 that is equal to 97.5.
Now, let us discuss a very important term percentage yield. What is Percentage Yield? Sometimes the reaction may not get completed according to the initial amount taken. Means whatever amount you have taken, whatever you predict it doesn’t happen. What are the possible reasons? The first possible reason is establishment of equilibrium. Now, this chapter you will read later in the portion of chemistry but let me tell you, that the amount of reactant you take, after a particular time it stabilises, it doesn’t go ahead. So, that is called as equilibrium. And therefore when you expect the corresponding product after the entire reaction, it doesn’t happen. Now, for the second reason, some of the reactants are not able to react due to formation of a protective layer. Suppose, there is a solid and over it we pour some liquid that reacts with the surface and then covers it up. Now, the remaining solid cannot react. And therefore we will not get the actual amount that we are thinking. Now, the third possible case would be when we take gaseous reactants they may simply escape out of the reaction chamber and will not give you the products. And the fourth possible reason can be other reactions happening in parallel for the same reactants. Like, till now you have studied N2 and H2 can give you ammonia, NH3, but they can also combine to give you N2H4, which is called as Hydrazine. So, you may not get ammonia in total that you are thinking. Now, in such cases the efficiency of the process is measured using percentage yield. What is percentage yield? It is the product mass that you actually get divided by the product mass that you theoretically predict times 100. Now, let us take an example for this.
A on controlled oxidation gives X according to the reaction. Stoichiometric is given to us. They are saying some of the reactant A oxidizes and only 70% of maximum yield is obtained. What mass of X is produced if 200 grams of A is taken? Now, what should be your strategy? First of all, we will calculate it normally. We will think that it is 100% yield and let’s find out how much mass of A that we can produce and then using the mass of X that we have calculated, we will calculate the actual mass by using the percentage yield 70%. So, let us see how we do it? We write down the stoichiometry of the reaction. 2A plus 9B gives you 2X plus 4Y plus 5Z. But what is the interesting part for me? 2A and 2X, so we write down. 2 mole A is going to produce 2 mole X. So, 1 mole A is going to produce 1 mole X and we first convert the moles into mass. So, 128 grams of A will give you 128 grams of X based on the molecule data given to us. So, from here we can say 128 grams of A produces 148 grams of X and therefore 1 gram of A is going to produce 148 by 128 grams of X. By taking 100% yield, 200 grams of A is going to produce 148 by 128 times 200 grams of X. Now, since percentage yield is equal to product mass actual divided by product mass theoretical times 100. We will rearrange this equation because we want mass actual. We can write it down as mass actual is equal to percentage yield by 100 times mass theoretical. Now, putting in the value, yield value is 70%, so, 70 by 100 times this much mass and that is going to give us 161.875 grams. So, first we calculate the actual amount or rather the theoretical amount and then we calculate the actual amount using the yield.
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