Mixed Problems: Set 2 (Stoichiometry and Gases)

Transcript

0:00Introduction

Hello everyone! So, one small puzzle before we start. I take four grams of hydrogen and sixteen grams of oxygen, and I explode them together. Four plus sixteen is twenty grams. So how much water do I get? Twenty grams, right? Mass is conserved, isn't it? Oh, but the answer is only eighteen grams of water! So where did the other two grams go? Today we will solve this, and a whole set of problems like it. And look, this unit gives one or two questions in almost every JEE Main shift, very often as numerical value questions. Each one should take you just one to two minutes. So, let's begin.

0:45One road for every problem

So, before any problem, one question. You read a stoichiometry question. What is the very first thing you write down? Yes, moles! The balanced equation never talks in grams. It talks in moles. So, grams into moles first: n equals small m by capital M, the mass divided by the molar mass, okay? The little arrow on screen just means, into. And if the question gives you a gas at STP? Then n equals the volume, V, divided by twenty two point four litres per mole, the molar volume these problems give you. Clear? Then cross the bridge: moles of one substance into moles of another, using the coefficients of the balanced equation. And finally, multiply back into grams, or litres. And that's it. That is the whole road. Quick test. Zero point six grams of carbon burns completely. Divide by twelve: zero point zero five moles of carbon. One carbon gives one carbon dioxide, so zero point zero five moles of carbon dioxide. Times forty four: two point two grams. Really simple, right?

1:56Limiting reagent: who runs out first?

Alright, now back to our puzzle. Sixteen grams of oxygen, and only four grams of hydrogen. So hydrogen runs out first, right? It is so much less. No, no, careful! This is the JEE trap. The limiting reagent is not simply the reactant present in smaller mass, or in smaller moles. See, you must bring in the balanced equation. Let's see how. So, first, the road: moles. Two moles of hydrogen react with one mole of oxygen, to give two moles of water. Hydrogen: four by two, two moles. Oxygen: sixteen by thirty two, zero point five moles. Now the topper's trick. Divide the moles of each reactant by its coefficient in the equation. Hydrogen: two by two, that is one. Oxygen: zero point five by one, that is zero point five. The smaller value is the limiting reagent. So oxygen limits, not hydrogen! And that smaller number, zero point five, is how many times the reaction runs, okay? Every product is its coefficient times that number. Water: two times zero point five, one mole. One mole of water is eighteen grams. And hydrogen? The reaction used two times zero point five, that is one mole. Two moles minus one: one mole left over, which is two grams of hydrogen. So, eighteen grams of water, plus two grams of leftover hydrogen: twenty grams. Nothing was lost! Those two grams were just extra hydrogen, with no oxygen left to react with. Mystery solved. Got it?

3:38Yield: actual over theoretical

Now, one more question for you. Our equation says how much product we should get. But in the lab, do we always get that full amount? Not always, right? Some is lost, some reacts another way. So we measure how much we actually got. Percentage yield equals the actual yield, divided by the theoretical yield, times one hundred. The little percent sign on screen is just that, percentage. And the theoretical yield is always based on the limiting reagent. Clear? One from the set. Twelve point four grams of phosphorus, P four, burns to give P four O ten. The molar mass of P four is one hundred twenty four, so twelve point four grams is zero point one moles of P four. One P four gives one P four O ten, so in theory, zero point one times two hundred eighty four: twenty eight point four grams. But we actually got only twenty five grams. Twenty five by twenty eight point four, times one hundred: eighty eight percent yield. Pretty straightforward, isn't it? Now another trap. Carbon is converted into carbon monoxide, and then into carbon dioxide, each step with ninety percent yield. Ten percent lost, then ten percent lost again, so eighty percent overall, right? Oh no, careful! For a chain of steps, multiply the yields. One mole, times zero point nine, times zero point nine: zero point eight one moles. Eighty one, not eighty. The second step only gets ninety percent of what the first step made. With me?

5:20Example 1: O2 volume at STP to oxidise 2.7 g of Al

Now, example one, an easy one. What volume of oxygen at STP, taking twenty two point four litres per mole, is needed to convert two point seven grams of aluminium into aluminium oxide? Pause the video, and try it first. Done? Okay, let's check. First, the balanced equation. Four aluminium plus three oxygen gives two aluminium oxide. Step one of our road: moles. Two point seven by twenty seven: zero point one moles of aluminium. Now the bridge. Four aluminium need three oxygen. So, do we multiply by three by four, or by four by three? Three by four, exactly! The coefficient of what you want, oxygen, goes on top; the coefficient of what you have, aluminium, goes below. Zero point one times three by four: zero point zero seven five moles of oxygen. Last step, back to litres. Zero point zero seven five times twenty two point four: one point six eight litres. Well under one minute. Nice and easy. And if you got this right, well done, you are on the right road!

6:33Example 2: percentage of KClO3 in a mixture with MnO2

Now a harder one, JEE level. Read it with me. Two point four five grams of a mixture of potassium chlorate and manganese dioxide, the catalyst, gives four hundred forty eight millilitres of oxygen at STP on heating. Find the percentage of potassium chlorate, to the nearest integer. Pause and try. Hint: first ask, who actually gives the oxygen? Done? Okay, let's check. Manganese dioxide is only the catalyst here, so all the oxygen comes from potassium chlorate. Two potassium chlorate give two potassium chloride, plus three oxygen. Now, moles of oxygen. Careful with units: four hundred forty eight millilitres is zero point four four eight litres. Divided by twenty two point four: zero point zero two moles. Now the bridge. Three oxygen come from two chlorate. So three by two, times zero point zero two: zero point zero three moles of chlorate, right? No, no, careful! This is the trap. That much chlorate would weigh more than three and a half grams, more than the whole mixture! We want chlorate, so its coefficient, two, goes on top, and oxygen's three goes below. Two by three, times zero point zero two: zero point zero one three three moles of potassium chlorate. Now its mass. The molar mass of potassium chlorate is one hundred twenty two point five. Topper's tip: do not round in the middle. Two by three, times zero point zero two, times one hundred twenty two point five: one point six three three grams. Percentage: one point six three three by two point four five, times one hundred. Sixty six point seven percent, and to the nearest integer, sixty seven. See? Moles, bridge, back to grams. The same road again.

8:33Quick revision: moles route, limiting reagent, yield

So, quick revision. One: moles first, cross over with the coefficients of the equation, then convert back. Two: the limiting reagent has the smallest moles divided by coefficient, not the smallest mass. Three: yield is actual over theoretical, times one hundred, and for a chain of steps, multiply the yields. Now go and try the practice problems on this page. Set a timer, one to two minutes each. You will find them easy, I promise. All the best!