Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You burn six grams of carbon in eight grams of oxygen. What do you get? Carbon dioxide, right? Carbon burns, carbon dioxide forms, simple. Wait, not so fast! The correct answer is carbon monoxide, and only carbon monoxide. Why? We will find out by the end of this class. This is our mixed problem class for JEE Main: the tools, the traps and the shortcuts that this whole unit is built on. So, let's begin.

0:37Counting particles: compare moles, not grams

So, first tool. Which has the most molecules: sixteen grams of oxygen, sixteen grams of nitrogen dioxide, seven grams of nitrogen, or two grams of hydrogen? One of the sixteen gram ones, right? Biggest mass, most molecules? No, no, careful! This is the first JEE trap. Molecules are counted by moles, not by grams. So divide each mass by its molar mass. Oxygen gas is thirty two, so sixteen by thirty two is zero point five mole. Nitrogen dioxide is forty six, so sixteen by forty six is about zero point three four eight mole. Nitrogen gas: seven by twenty eight, zero point two five mole. And hydrogen gas: two by two, one full mole! Most moles means most molecules; the double arrow on screen is read as, implies. The answer is option d, just two grams of hydrogen! Light molecules, so many more of them. Clear? And if they ask for atoms, not molecules? Take zero point one mole of a triatomic gas: three atoms in every molecule. Moles, times three, times Avogadro's number, six point zero two two times ten to the power twenty three. That gives one point eight zero six times ten to the power twenty three atoms. Look, that times three is the step students forget.

2:04Count the parts inside one formula unit

Next tool. How many moles of ions are present in one hundred millilitres of zero point two molar potassium ferrocyanide? Its formula is on screen: K four, F e, C N six. First tell me, how many ions does one formula unit give? Hmm, keep your answer ready; we will check it in a moment. Alright, first the salt itself. One hundred millilitres is zero point one litre. Times zero point two molar: zero point zero two mole of the salt. Now break one formula unit. Four potassium ions, and one complex ion, the whole bracket, with a charge of four minus. Careful: that bracket counts as one ion only. So five ions each, and zero point zero two times five is zero point one mole of ions. Same tool, the other way round. What mass of calcium phosphate contains six point two grams of phosphorus? Its molar mass, in brackets, is three hundred and ten, and phosphorus is thirty one. So moles of phosphorus, written n of P on screen: six point two by thirty one, zero point two. But see, each formula unit has two phosphorus atoms. So the salt is only half of that: zero point one mole. Zero point one times three hundred and ten: thirty one grams. So always ask one question: how many of this part sit in one formula unit? Got it?

3:33Limiting reagent: the smaller supply decides

Third tool, and JEE loves it. Zero point five mole of barium chloride is mixed with zero point two mole of sodium sulphate. Find the mass of barium sulphate precipitated; its molar mass is two hundred and thirty three. So which number decides the answer? Many students grab the first number, zero point five, and multiply. No, no! You cannot make more barium sulphate than your sulphate allows, right? The reaction is one to one, okay? Barium chloride, zero point five; sodium sulphate, only zero point two. The sign on screen is read, less than. So the sulphate runs out first: it is the limiting reagent, and the extra barium chloride is simply left over. So barium sulphate is zero point two mole. Zero point two times two hundred and thirty three: forty six point six grams. Remember: the reagent that runs out first decides the product. Every single time.

4:34Example 1: volume of CaCl2 to precipitate all the silver

Right, let's do example one. What volume of zero point zero five molar calcium chloride precipitates all the silver from fifty millilitres of zero point one molar silver nitrate? Pause the video, and try it first. Done? Okay, let's check. Work in millimoles: millilitres times molarity. Fifty times zero point one: five millimoles of silver ions. Each silver ion takes one chloride ion, so we need five millimoles of chloride. Now, how much calcium chloride is that? Five millimoles, right? Hold on, careful! Look at its formula once more. Each calcium chloride gives two chloride ions: our second tool, remember? So we need only half: two point five millimoles of calcium chloride. Volume is millimoles over molarity: two point five over zero point zero five. Fifty millilitres. Forget that two, and your answer comes out double. That is the trap.

5:38Example 2: 6 g carbon burnt in 8 g oxygen, CO or CO2

Now the harder one, and it is our opening puzzle. Six grams of carbon is burnt in eight grams of oxygen. Carbon monoxide and carbon dioxide may both form. What is the product? Pause the video and try it. Hint: grams will fool you, so count moles first. Done? Okay, let's check. Carbon: six by twelve, zero point five mole. Oxygen gas: eight by thirty two, zero point two five mole. To make carbon dioxide, each carbon atom needs one whole oxygen molecule. So zero point five mole of carbon needs zero point five mole of oxygen gas. Do we have that much? No! Only zero point two five, just half. But carbon monoxide needs only one oxygen atom per carbon, so one oxygen molecule serves two carbon atoms. Zero point five mole of carbon needs just zero point two five mole of oxygen gas. And that is exactly what we have! So all the carbon becomes carbon monoxide: zero point five mole. Its molar mass is twenty eight, so fourteen grams of carbon monoxide. And see, six grams plus eight grams is fourteen grams: every gram is used. Puzzle solved! The oxygen ran out, and it decided the product. All clear?

7:04Quick revision: moles, ions per formula, limiting reagent

So, quick revision. One: to compare numbers of particles, compare moles, never grams. Two: for atoms or ions inside a formula, multiply by how many sit in one formula unit. Three: when two reactants are mixed, find the limiting reagent first; it decides the product. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!