Reading a Balanced Equation

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take one litre of nitrogen gas and three litres of hydrogen gas, at the same temperature and pressure. Suppose they react completely to make ammonia. How many litres of ammonia do you get? Four, right? One plus three is four. No! You get only two litres. So then, where did the other two litres go? By the end of this video, this one plus three equals two will make perfect sense. And look, in JEE, every stoichiometry question starts with reading a balanced equation. So, let's begin.

0:41A balanced equation is a statement about moles

So, here is the equation for making ammonia. N two, plus three H two, gives two N H three. The little g in brackets means a gas, okay? Now look at the numbers in front: one, three and two. No number in front of N two simply means one. These are the coefficients. Now tell me, what do these coefficients actually count? Grams? No, no. Not grams. Yes, they count moles! So, read the equation like a sentence. One mole of N two reacts with three moles of H two, to give two moles of N H three. So a balanced equation is really a statement about moles. Clear? Alright, now back to our puzzle. One plus three, four moles go in, and only two moles come out. So, has some mass been lost? One mole of N two is twenty eight grams. Three moles of H two is six grams. Twenty eight plus six: thirty four grams go in. And two moles of ammonia? Exactly thirty four grams. See, not even one gram is lost! And for gases at the same temperature and pressure, the same numbers work for volumes. One volume of nitrogen, plus three volumes of hydrogen, give two volumes of ammonia. So, puzzle solved: one litre plus three litres gives two litres. Nothing vanished. The mass is the same; only the number of moles went down.

2:13The stoichiometric ratio

Now, let's make it general. Take any balanced equation: small a moles of A, plus small b moles of B, give small c moles of C, plus small d moles of D, okay? The small letters are the coefficients. Quick question. In our ammonia equation, for one mole of nitrogen, how many moles of hydrogen react? Three, correct! The moles always march in step with the coefficients. So here is the key result, the stoichiometric ratio. Here n stands for the number of moles. n of A over a, equals n of B over b, equals n of C over c, equals n of D over d. Divide the moles of each substance by its own coefficient, and you get the same number for all. Write this down. Got it? But look, one small condition. These are the moles that actually react, or actually form. Any extra reactant left over in the flask does not count, okay? Alright, now look at this equation. Two H two, plus O two, gives two H two O, okay? Mass is conserved in every reaction, we just saw that. So, are the total moles also conserved? Count them. Two plus one: three moles go in. And only two moles come out. So, mass is conserved, but moles need not be. Never expect the moles on both sides to match. That is a classic JEE trap, got it?

3:44Topper's trick: balance by inspection

Now, all of this works only when the equation is balanced. So, how do toppers balance quickly? Step one: start with the element that appears in only one compound on each side. Step two: leave hydrogen and oxygen, and free elements like O two, for the end. Why? Because hydrogen and oxygen usually sit in many compounds, while O two contains only oxygen, so you can change it without disturbing any other element, right? Step three: if a fraction appears, don't panic. Clear it at the end, by multiplying the whole equation through. Okay? So, let's try it on butane burning in oxygen. C four H ten, plus O two, gives C O two, plus H two O, okay? Which element first? Carbon, exactly! It sits in only one compound on each side. Four carbons on the left, so four C O two on the right. Ten hydrogens, so five H two O, okay? Now oxygen, last of all: eight oxygen atoms from the C O two, and five from the water. Thirteen. So we need thirteen by two O two. Oh, a fraction! Simple, just multiply everything by two. Two C four H ten, plus thirteen O two, gives eight C O two, plus ten H two O, okay? Perfectly balanced.

5:13Example 1: grams of oxygen to burn 16 g of methane

Alright, time for the first example. How many grams of oxygen are needed to burn sixteen grams of methane completely? The equation is given: C H four, plus two O two, gives C O two, plus two H two O, okay? Pause the video, and try it first. Done? Okay, let's check. Some students do this: a two in front of O two, so two times sixteen, thirty two grams of oxygen, right? No, no, careful! The coefficients count moles, not grams. Never multiply grams by a coefficient. So, first go to moles. The molar mass of methane is twelve plus four, that is sixteen. Sixteen grams over sixteen: exactly one mole of methane. Now the stoichiometric ratio. Moles of O two over two, equals moles of methane over one. So the moles of O two are two times one: two moles. And now back to grams. One mole of O two is thirty two grams, so two moles is sixty four grams. Sixty four grams, not thirty two. That shortcut gives exactly half the answer. Clear?

6:26Example 2: mass of KClO3 that gives 9.6 g of oxygen

Now a slightly harder one, JEE level. What mass of potassium chlorate, K C l O three, must be decomposed to give nine point six grams of oxygen? Hint: balance the equation first. Pause the video, and try it. Done? Okay, let's check. Potassium chlorate breaks into potassium chloride, K C l, and oxygen. Potassium and chlorine are already fine. Oxygen is the free element, so it comes last, just like our trick. Three oxygens on the left, and O two, two oxygens, on the right, okay? How do we balance it? Yes, two K C l O three gives two K C l, plus three O two. Six oxygen atoms on each side. Completely balanced. Now go to moles. Nine point six grams of oxygen, divided by thirty two: zero point three moles of O two. Now the ratio. Do we multiply zero point three by three by two, or by two by three? Many students write three by two here, and get zero point four five moles. Wrong! That is the trap. Use the stoichiometric ratio. Moles of potassium chlorate over two, equals moles of O two over three. So that is two thirds of zero point three: zero point two moles. Topper's check: potassium chlorate has the smaller coefficient, two against three, so it must have fewer moles. Zero point two is less than zero point three. Correct! Last step, back to grams. The molar mass of potassium chlorate is one hundred twenty two point five. Zero point two times one hundred twenty two point five: twenty four point five grams. The wrong ratio would give fifty five point one two five grams, far too much. With me?

8:17Quick revision: coefficients count moles, not grams

So, quick revision. One: coefficients count moles, and for gases at the same temperature and pressure, volumes too. Never grams. Two: moles divided by the coefficient is the same number for every substance. Three: mass is always conserved, but moles need not be. Remember our one plus three equals two! Now go and try the practice problems on this page. Go to moles, use the ratio, and they will feel easy. All the best!