The Limiting Reagent

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. I take ten grams of hydrogen and sixty four grams of oxygen, and I burn them to make water. Ten grams against sixty four grams. Hydrogen is so little, isn't it? So which gas will run out first? Hydrogen, obviously, right? Well, at the end of the reaction, it is hydrogen that is left over! Wait, how is that possible? Today we will find out. The idea is called the limiting reagent. And look, in JEE, so many stoichiometry numericals quietly hide this one question: which reactant finishes first? So, let's get started.

0:42The limiting reagent

So, first, a quick recap. Here is the balanced equation for ammonia: N two plus three H two gives two N H three. It says, one mole of nitrogen reacts with three moles of hydrogen, always in this one to three ratio. This fixed ratio from the balanced equation is the stoichiometric ratio, okay? But in a question, nobody promises you this ratio. Suppose the amounts we mix are not in the ratio one to three. What happens then? Yes, one of them gets used up first, and some of the other one is still sitting there. Exactly. And that reactant gets a special name. The reactant that is used up first is called the limiting reagent. Limiting, because it puts a limit on the reaction. Got it? Now think. Once the limiting reagent is finished, can any more product form? No, it can't! The other reactant has nothing left to react with. So the limiting reagent decides the amount of product. Adding more of the other reactant does not help at all. Clear? And the other reactants? They are in excess. Part of them reacts, and the extra is simply left over at the end. In questions you will see the words reagent in excess, or reagent left over. Same thing, okay?

2:00JEE trap: is the smaller amount limiting?

So, how do we find the limiting reagent? Most students say: easy, whichever reactant is present in the smaller amount finishes first. Is that right? Hmm, careful. This is the classic JEE trap. The limiting reagent is not simply the reactant with the smaller mass, or the smaller number of moles. Why not? Let's see. Look at ammonia again. For every one mole of nitrogen, the reaction eats three moles of hydrogen. So hydrogen is used up three times as fast! Even if you start with more moles of hydrogen, it can still finish first. With me? So the amount alone tells you nothing. You must compare it with what the reaction needs, and that is the coefficient in the balanced equation. Always divide by the stoichiometric coefficient. Write this down.

2:51Topper's trick: divide moles by the coefficient

Now let me share a trick that toppers use. It works for any reaction. Take a general reaction: small a moles of reactant capital A, plus small b moles of reactant capital B, give the products, okay? The dots on screen just mean whatever products are formed. Step one: for each reactant, divide its moles by its own coefficient. So, n sub A over small a, and n sub B over small b. Here n sub A is the number of moles of A that you actually have, okay? Now tell me. Which one is the limiting reagent: the reactant with the smaller value, or with the larger value? The smaller one, exactly! The smaller value gives the limiting reagent. It is the reactant that gets exhausted first, once you account for how fast the reaction uses it up. And here is the beautiful part. That smaller value is the number of times the reaction runs. If it is two, the whole balanced equation happens exactly two times, and then the limiting reagent is finished. So every product is just its coefficient, times that number. The same goes for the reactant in excess: the amount used is its coefficient times that number, and whatever is not used is left over. That's it, one division and one multiplication. Really fast!

4:14Example 1: limiting reagent and moles of ammonia formed

Now, let's try one. Three moles of hydrogen and two moles of nitrogen react to form ammonia. Find the limiting reagent, and the moles of ammonia formed. Pause the video, and try it first. Done? Okay, let's check. Nitrogen is two moles, hydrogen is three moles. Nitrogen is less, so nitrogen is the limiting reagent, right? No, no, careful! That is exactly the trap we just saw. First, always write the balanced equation. N two plus three H two gives two N H three. Hydrogen has the coefficient three, and nitrogen has one, okay? Now divide. Hydrogen: three moles divided by three, that is one. Nitrogen: two moles divided by one, that is two. The smaller value is one, and it belongs to hydrogen. So hydrogen is the limiting reagent, even though we have more moles of it! And the reaction runs exactly one time. So the ammonia formed is its coefficient, two, times one. Two moles of ammonia. And nitrogen? Its coefficient is one, so the reaction used one times one, that is one mole. Two minus one: one mole of nitrogen is left over, in excess. See how the trick gives everything in one go? Clear?

5:37Example 2: water from 10 g of H2 and 64 g of O2

Now a JEE pattern question, and it is our puzzle from the start. Ten grams of hydrogen and sixty four grams of oxygen are ignited to form water. Find the mass of water formed, and the gas left over. Pause and try. Hint: change grams to moles first. Done? Okay, let's check. Ten grams is much less than sixty four grams, so hydrogen runs out first, right? Oh no, the same trap again, just with grams this time! Comparing masses tells us nothing. Step one, grams to moles. Hydrogen gas, H two, has a molar mass of two grams per mole, so n of H two is ten by two, five moles. Oxygen gas, O two, has thirty two, so sixty four by thirty two gives two moles. Now the balanced equation: two H two plus O two gives two H two O, okay? Hydrogen has the coefficient two, and oxygen has one. Now divide each by its coefficient. Hydrogen: five by two, that is two point five. Oxygen: two by one, that is two. So which one limits? Think for a second. Got it? Let's see. Oxygen, yes! Two is smaller than two point five. So oxygen is the limiting reagent, and the reaction runs two times. Water has the coefficient two, so the water formed is two times two, four moles. One mole of water is eighteen grams, so four times eighteen: seventy two grams of water. Hydrogen used is its coefficient, two, times two, that is four moles. We had five, so one mole is left over, and one mole of hydrogen gas is two grams. And there you go, puzzle solved! Those ten grams of hydrogen looked small, but they are five moles, and oxygen needs only four of them. So hydrogen is the reagent in excess, with two grams left over. And one quick check that toppers always do. Before: ten plus sixty four, that is seventy four grams. After: seventy two grams of water plus two grams of hydrogen, also seventy four grams. Mass is conserved, so everything adds up. Got it?

7:59Quick revision: divide moles by the coefficient

So, quick revision. The limiting reagent is the reactant used up first, and it fixes the amount of product. To find it, divide the moles of each reactant by its coefficient. The smaller value limits, and it tells you how many times the reaction runs. And never judge by the smaller mass or the smaller moles alone. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!